TREE2.cpp

/*
Preorder traversal (Iterative) 
Medium Accuracy: 71.48% Submissions: 12378 Points: 4
Given a binary tree. Find the preorder traversal of the tree without using recursion.


Example 1:

Input:
           1
         /   \
        2     3
      /  \
     4    5
Output: 1 2 4 5 3
Explanation:
Preorder traversal (Root->Left->Right) of 
the tree is 1 2 4 5 3.
*/
 vector<int> preOrder(Node* root)
    {
    vector<int>v;
    if(root==NULL)return v;
    stack<Node*>s;
    s.push(root);
    while(!s.empty()){
        Node*temp=s.top();
        s.pop();
        v.push_back(temp->data);
        if(temp->right){
            s.push(temp->right);
        }
        if(temp->left){
            s.push(temp->left);
        }
    }
    return v;
 }
/*
Inorder Traversal (Iterative) 
Medium Accuracy: 74.94% Submissions: 11060 Points: 4
Given a binary tree. Find the inorder traversal of the tree without using recursion.

Example 1

Input:
           1
         /    \
       2       3
      /   \
    4     5
Output: 4 2 5 1 3
Explanation:
Inorder traversal (Left->Root->Right) of 
the tree is 4 2 5 1 3.
*/
vector<int> inOrder(Node* root)
    {
     vector<int>v;
    if(root==NULL)return v;
    stack<Node*>s;
    Node*temp=root;
    while(!s.empty()|| temp!=NULL){
        while(temp!=NULL){
            s.push(temp);
            temp=temp->left;
        }
        temp=s.top();
        s.pop();
        v.push_back(temp->data);
        temp=temp->right;    
        //code here
    }
    return v;
    }
/*
Postorder Traversal (Iterative) 
Medium Accuracy: 76.51% Submissions: 7228 Points: 4
Given a binary tree. Find the postorder traversal of the tree without using recursion.

Example 1

Input:
           1
         /   \
        2     3
      /  \
     4    5

Output: 4 5 2 3 1
Explanation:
Postorder traversal (Left->Right->Root) of 
the tree is 4 5 2 3 1.
*/
 vector<int> postOrder(Node* node) {
        stack<Node*> s;
vector<int> res;
s.push(node);
while(!s.empty())
{
node=s.top();
s.pop();
res.push_back(node->data);
if(node->left!=NULL)
s.push(node->left);
if(node->right!=NULL)
s.push(node->right);

}
reverse(res.begin(),res.end());
return res;
    }
/*
Make Binary Tree From Linked List 
Medium Accuracy: 51.16% Submissions: 21586 Points: 4
Given a Linked List Representation of Complete Binary Tree. The task is to construct the Binary tree.
Note : The complete binary tree is represented as a linked list in a way where if root node is stored at position i, its left, and right children are stored at position
2*i+1, 2*i+2 respectively.
 

*/
void convert(Node *head, TreeNode *&root) {
    if(head==NULL)return;
    root=new TreeNode(head->data);
    queue<TreeNode*>q;

    q.push(root);
    while(!q.empty()){
        TreeNode*temp=q.front();
        q.pop();
        if( head->next){
            temp->left=new TreeNode(head->next->data);
            q.push(temp->left);
            head=head->next;
        }
        if(head->next){
            temp->right=new TreeNode(head->next->data);
             q.push(temp->right);
            head=head->next;
        }
        
    }
    /*
    Count Number of SubTrees having given Sum 
Medium Accuracy: 50.13% Submissions: 25893 Points: 4
Given a binary tree and an integer X. Your task is to complete the function countSubtreesWithSumX() that returns the count of the number of subtress having 
total node’s data sum equal to the value X.
Example: For the tree given below:            

              5
            /    \
        -10     3
        /    \    /  \
      9     8  -4 7

Subtree with sum 7:
             -10
            /      \
          9        8

and one node 7.
    */
  int solve(Node*root,int &count,int X){
    if(root==NULL)return 0;
    int l=solve(root->left,count,X);
    int r=solve(root->right,count,X);
    int sum=l+r+root->data;
    if(sum==X)count++;
    return sum;
}
int countSubtreesWithSumX(Node* root, int X)
{
    int count=0;
    solve(root,count,X);
    return count;

}
/*
Leftmost and rightmost nodes of binary tree 
Medium Accuracy: 50.0% Submissions: 14212 Points: 4
Given a Binary Tree of size N, Print the corner nodes ie- the node at the leftmost and rightmost of each level.


Example 1:

Input :
         1
       /  \
     2      3
    / \    / \
   4   5  6   7    
Output: 1 2 3 4 7
Explanation:
Corners at level 0: 1
Corners at level 1: 2 3
Corners at level 2: 4 7

*/  
 void printCorner(Node *root)
{
    if(!root){
        return;
    }
    vector<int>v;
    queue<Node*>q;
    q.push(root);
    while(!q.empty()){
        int size=q.size();
         int n=size;
       
        while(size--){
            
            Node*temp=q.front();
            q.pop();
            if(n==1){
                 v.push_back(temp->data);
            }
           if(n>1){
               if(size==0 || size==n-1){
                    v.push_back(temp->data);
                    
               }
               
           }
            if(temp->left){
                q.push(temp->left);
                
            }
            if(temp->right){
                q.push(temp->right);             
            }      
        }
    }
for(int i=0;i<v.size();i++){
    cout<<v[i]<<" ";
}   
 /*
 Diagonal Traversal of Binary Tree 
Medium Accuracy: 49.87% Submissions: 31020 Points: 4
Given a Binary Tree, print the diagonal traversal of the binary tree.

Consider lines of slope -1 passing between nodes. Given a Binary Tree, print all diagonal elements in a binary tree belonging to same line.

Example 1:

Input :
            8
         /     \
        3      10
      /   \      \
     1     6     14
         /   \   /
        4     7 13
Output : 8 10 14 3 6 7 13 1 4
 */   
 vector<int> diagonal(Node *root)
{
      vector<int> ans;
if(root==NULL)return ans;
queue<Node*>q;
q.push(root);

while(!q.empty()){

Node* temp = q.front();
q.pop();
while(temp!=NULL){
ans.push_back(temp->data);
if(temp->left)q.push(temp->left);
temp=temp->right;
}}
return ans;
   // your code here
}
/*
Bottom View of Binary Tree 
Medium Accuracy: 45.32% Submissions: 91064 Points: 4
Given a binary tree, print the bottom view from left to right.
A node is included in bottom view if it can be seen when we look at the tree from bottom.

                      20
                    /    \
                  8       22
                /   \        \
              5      3       25
                    /   \      
                  10    14

For the above tree, the bottom view is 5 10 3 14 25.
If there are multiple bottom-most nodes for a horizontal distance from root, then print the later one in level traversal. For example, in the below diagram, 3 and 4 are both the bottommost nodes at horizontal distance 0, we need to print 4.

                      20
                    /    \
                  8       22
                /   \     /   \
              5      3 4     25
                     /    \      
                 10       14

For the above tree the output should be 5 10 4 14 25.
*/        
  vector <int> bottomView(Node *root) {
         vector<int>v;
        map<int,int>m;
        if(root==NULL)return v;
        queue<pair<Node*,int>>q;
        q.push({root,0});
         while(!q.empty()){
          
                 Node*temp=q.front().first;
                 int h=q.front().second;
                 q.pop();
                
                     m[h]=temp->data;
                 
                 if(temp->left){
                    q.push({temp->left,h-1}) ;
                 }
                 if(temp->right){
                     q.push({temp->right,h+1});
                 }
             }
         
        for(auto x:m){
            v.push_back(x.second);
        }
      return v;   
     
    }   
           /*
           Vertical Traversal of Binary Tree 
Medium Accuracy: 32.45% Submissions: 100k+ Points: 4
Given a Binary Tree, find the vertical traversal of it starting from the leftmost level to the rightmost level.
If there are multiple nodes passing through a vertical line, then they should be printed as they appear in level order traversal of the tree.

Example 1:

Input:
           1
         /   \
       2       3
     /   \   /   \
   4      5 6      7
              \      \
               8      9           
Output: 
4 2 1 5 6 3 8 7 9 
           */ 
  vector<int> verticalOrder(Node *root)
    {
         vector<int>v;
        if(!root)return v;
        queue<pair<Node*,int>>q;
        map<int,vector<int>>m;
        q.push({root,0});
        while(!q.empty()){
            pair<Node*,int>temp=q.front();
            q.pop();
            Node*c=temp.first;
            int h=temp.second;
            m[h].push_back(c->data);
            if(c->left)q.push({c->left,h-1});
            if(c->right)q.push({c->right,h+1});
            
        }
        for(auto x:m){
            vector<int>l=x.second;
            for(auto y:l){
                v.push_back(y);
            }
        }
        return v;
        //Your code here
    }
  /*
  Check if subtree 
Medium Accuracy: 43.66% Submissions: 79673 Points: 4
Given two binary trees with head reference as T and S having at most N nodes. The task is to check if S is present as subtree in T.
A subtree of a tree T1 is a tree T2 consisting of a node in T1 and all of its descendants in T1.

Example 1:

Input:
T:      1          S:   3
      /   \            /
     2     3          4
   /  \    /
  N    N  4
Output: 1 
Explanation: S is present in T
  */   
  bool solve(Node*t,Node*s){
     if(!t and !s)return true;
     if(!t || !s)return false;
     if(t->data!=s->data){
         return false;
     }
     if(solve(t->left,s->left)&& solve(t->right,s->right)){
         return true;
     }
     return false;
 }
    //Function to check if S is a subtree of tree T.
    bool isSubTree(Node* T, Node* S) 
    {
        if(!T  and !S){
            return true;
        }
        if(T!=NULL){
        if(isSubTree(T->left,S)|| isSubTree(T->right,S)||solve(T,S))return true;
        }
        return  false;
  
    }
   /*
   Construct Binary Tree from Parent Array 
Medium Accuracy: 45.96% Submissions: 23829 Points: 4
Given an array of size N that can be used to represents a tree. The array indexes are values in tree nodes and array values give the parent node of that particular index (or node). The value of the root node index would always be -1 as there is no parent for root. Construct the standard linked representation of Binary Tree from this array representation.


Example 1:

Input:
N = 7
parent[] = {-1,0,0,1,1,3,5}
Output: 0 1 2 3 4 5 6
Explanation: the tree generated
will have a structure like 
          0
        /   \
       1     2
      / \
     3   4
    /
   5
 /
6
   
   */         
Node *createTree(int parent[], int N)
    {
        unordered_map<int,Node*>mp;
        for(int i=0;i<N;i++){
            Node*temp=new Node(i);
            mp[i]=temp;
        }
        Node*root;
        Node*rooot;
        for(int i=0;i<N;i++){
            if(parent[i]==-1){
                rooot=mp[i];
                
            }
            else{
                Node*temp=mp[i];
                int val=parent[i];
                root=mp[val];
                if(!root->left){
                    root->left=temp;
                }
                else{
                    root->right=temp;
                }
            }
        }
        return rooot;
        // Your code here
    }
            /*
            Preorder Traversal and BST 
Medium Accuracy: 58.03% Submissions: 5042 Points: 4
Given an array arr[ ] of size N consisting of distinct integers, write a program that returns 1 if given array can represent preorder traversal of a possible BST, else returns 0.

Example 1:

Input:
N = 3
arr = {2, 4, 3}
Output: 1
Explaination: Given arr[] can represent
preorder traversal of following BST:
               2
                \
                 4
                /
               3
            */
     int canRepresentBST(int pre[], int n) {
         stack<int> s;
       int root = INT_MIN;
       for (int i=0; i<n; i++)
       {
           if (pre[i] < root)
               return 0;
   
           while (!s.empty() && s.top()<pre[i])
           {
               root = s.top();
               s.pop();
           }
           s.push(pre[i]);
       }
       return 1;
        // code here
    }
/*
traversal of BST. The task is to print its postorder traversal.

Example 1:

Input:
N = 5
arr[]  = {40,30,35,80,100}
Output: 35 30 100 80 40
Explanation: PreOrder: 40 30 35 80 100
InOrder: 30 35 40 80 100
Therefore, the BST will be:
              40
           /      \
         30       80
           \        \   
           35      100
Hence, the postOrder traversal will
be: 35 30 100 80 40
*/        
Node*solve(Node*&root,int x){
    if(root==NULL){
        return root=newNode(x);
    }
    if(root->data>x){
        root->left=solve(root->left,x);
    }
    else{
        root->right=solve(root->right,x);
    }
    return root;
}
Node* post_order(int pre[], int size)
{
    Node*root=NULL;
    for(int i=0;i<size;i++){
        solve(root,pre[i]);
    }
    return root;
   
}
 /*
 Clone a Binary Tree 
Medium Accuracy: 55.34% Submissions: 15640 Points: 4
Given a special binary tree having random pointers along with the usual left and right pointers. Clone the given tree.


Example 1:

Input:
 */           
 void insertmid(Node *root)
{
if(!root)
return;

Node*temp= new Node(root->data);
temp->left=root->left;
temp->right=root->right;
root->left=temp;

insertmid(temp->left);
insertmid(temp->right);
}
void random(Node *root)
{
if(!root)
return;

root->left->random=(root->random)?root->random->left:NULL;

random(root->left->left);
random(root->left->right);
}
void restore (Node *root)
{
if(!root)
return;

Node*copy=root->left;
root->left=copy->left;
root->right=copy->right;

copy->left=(root->left)?root->left->left:NULL;
copy->right=(root->right)?root->right->left:NULL;

restore(root->left);
restore(root->right);
}
Node* cloneTree(Node* root)
{
insertmid(root);
random(root);

Node* cloneroot=root->left;
restore(root);

return cloneroot;
}
/*
Construct Tree from Preorder Traversal 
Medium Accuracy: 50.68% Submissions: 13694 Points: 4
Construct a binary tree of size N using two given arrays pre[] and preLN[]. Array pre[] represents preorder traversal of a binary tree. Array preLN[] has only two possible values ‘L’ and ‘N’. The value ‘L’ in preLN[] indicates that the corresponding node in Binary Tree is a leaf node and value ‘N’ indicates that the corresponding node is a non-leaf node.
Note: Every node in the binary tree has either 0 or 2 children.

Example 1:

Input :      
N = 5
pre[] = {10, 30, 20, 5, 15}
preLN[] = {N, N, L, L, L}

Output:
          10
        /    \
      30      15
     /  \     
   20    5   
*/      
 struct Node *constructTree(int n, int pre[], char preLN[])
{
    Node*root=new Node(pre[0]);
    stack<Node*>s;
    s.push(root);
    int i=1;
    while(i<n){
        Node*temp=s.top();
        if(temp->left==NULL){
            temp->left=new Node(pre[i]);
            if(preLN[i]=='N'){
                s.push(temp->left);
                
            }
             i++;
        }
             else if(temp->right==NULL){
            temp->right=new Node(pre[i]);
            if(preLN[i]=='N'){
                s.push(temp->right);
            }
            i++;
        }
        
        else{
            s.pop();
        }

    }
   return root; 
 }
            /*
   Serialize and Deserialize a Binary Tree 
Medium Accuracy: 47.46% Submissions: 35481 Points: 4
Serialization is to store a tree in an array so that it can be later restored and Deserialization is reading tree back from the array. Now your task is to complete the function serialize which stores the tree into an array A[ ] and deSerialize which deserializes the array to the tree and returns it.
Note: The structure of the tree must be maintained. Multiple nodes can have the same data.

Example 1:

Input:
      1
    /   \
   2     3
Output: 2 1 3         
            
            */
    void solve(Node*root,vector<int>&v){
        if(root==NULL){
            v.push_back(-1);
            return;
        }
        v.push_back(root->data);
        solve(root->left,v);
        solve(root->right,v);
        
    }
    //Function to serialize a tree and return a list containing nodes of tree.
    vector<int> serialize(Node *root) 
    {vector<int>v;
    solve(root,v);
    return v;
        
        //Your code here
    }
    
    //Function to deserialize a list and construct the tree.
  Node* deSerializeUtil(vector<int> &A,int &pointer)
{
if(A[pointer]==-1){
pointer++;
return NULL;
}
Node *root = new Node(A[pointer]);
pointer++;
root->left = deSerializeUtil(A,pointer);
root->right = deSerializeUtil(A,pointer);

return root;
}

Node * deSerialize(vector<int> &A)
{
//Your code here
int pointer=0;
return deSerializeUtil(A,pointer);
} 
   /*
   Boundary Traversal of binary tree 
Medium Accuracy: 26.78% Submissions: 100k+ Points: 4
Given a Binary Tree, find its Boundary Traversal. The traversal should be in the following order: 

Left boundary nodes: defined as the path from the root to the left-most node ie- the leaf node you could reach when you always travel preferring the left subtree over
the right subtree. 
Leaf nodes: All the leaf nodes except for the ones that are part of left or right boundary.
Reverse right boundary nodes: defined as the path from the right-most node to the root. The right-most node is the leaf node you could reach when you always travel
preferring the right subtree over the left subtree. Exclude the root from this as it was already included in the traversal of left boundary nodes.
Note: If the root doesn't have a left subtree or right subtree, then the root itself is the left or right boundary. 

Example 1:

Input:
        1 
      /   \
     2     3  
    / \   / \ 
   4   5 6   7
      / \
     8   9
   
Output: 1 2 4 8 9 6 7 3
Explanation:
     
   */
            void leaf(Node*root,vector<int>&v){
    if(root==NULL){
        return;
    }
    if(root->left ==NULL && root->right==NULL){
        v.push_back(root->data);
    }
    leaf(root->left,v);
    leaf(root->right,v);
    
}
void fun(Node*root,vector<int>&v){
    if(root==NULL){
        return;
    }
    if(root->right){
        fun(root->right,v);
     v.push_back(root->data);   
    }
    else if(root->left){
    fun(root->left,v);
     v.push_back(root->data);    
    }
}
void solve(Node*root,vector<int>&v){
    if(root==NULL){
        return;
    }
    if(root->left){
        v.push_back(root->data);
        solve(root->left,v);
    }
    else if(root->right){
          v.push_back(root->data);
        solve(root->right,v);
    }
    
}
    vector <int> boundary(Node *root)
    {
        vector<int>v;
        if(root==NULL){
            return v;
        }
        v.push_back(root->data);
        solve(root->left,v);
        leaf(root,v);
        fun(root->right,v);
     
    }
 /*
 Min distance between two given nodes of a Binary Tree 
Medium Accuracy: 45.05% Submissions: 49448 Points: 4
Given a binary tree and two node values your task is to find the minimum distance between them.

Example 1:

Input:
        1
      /  \
     2    3
a = 2, b = 3
Output: 2
Explanation: The tree formed is:
       1
     /   \ 
    2     3
We need the distance between 2 and 3.
Being at node 2, we need to take two
steps ahead in order to reach node 3.
The path followed will be:
2 -> 1 -> 3. Hence, the result is 2. 
 */        
  int dis(Node*root,int a){
            if(root==NULL)return 0;
            if(root->data==a)return 1;
            int l=dis(root->left,a);
            int r=dis(root->right,a);
            if(!l && !r)return 0;
            return l+r+1;
    }
    Node*lca(Node*root,int a,int b){
        if(root==NULL){
            return root;
        }
        if(root->data==a || root->data==b){
            return root;
        }
        Node*l=lca(root->left,a,b);
        Node*r=lca(root->right,a,b);
        if(l&&r)return root;
        if(l){
            return l;
        }
        else{
            return r;
        }
    }
    int findDist(Node* root, int a, int b) {
         if(root){
          int x=dis(root,a);
          int y=dis(root,b);
          Node*temp=lca(root, a,b);
          int z=dis(root,temp->data);
          return (x+y)-(2*z);
        }
        // Your code he
/*
Construct Tree from Inorder & Preorder 
Medium Accuracy: 36.26% Submissions: 56170 Points: 4
Given 2 Arrays of Inorder and preorder traversal. Construct a tree and print the Postorder traversal. 

Example 1:

Input:
N = 4
inorder[] = {1 6 8 7}
preorder[] = {1 6 7 8}
Output: 8 7 6 1
*/
  int search(int *in,int x,int n){
        for(int i=0;i<n;i++){
            if(in[i]==x){
                return i;
            }
        }
        return -1;
    }
    Node* buildTree(int in[],int pre[], int n)
    {
     if(n==0)return NULL;
    int i=search(in,pre[0],n);
    Node*root=new Node(pre[0]);
    root->left=buildTree(in,pre+1,i);
    root->right=buildTree(in+i+1,pre+i+1,n-i-1);
    return root;
    }
      / *
        Tree from Postorder and Inorder 
Medium Accuracy: 49.98% Submissions: 24325 Points: 4
Given inorder and postorder traversals of a Binary Tree in the arrays in[] and post[] respectively. The task is to construct the binary tree from these traversals.

 

Example 1:

Input:
N = 8
in[] = 4 8 2 5 1 6 3 7
post[] =8 4 5 2 6 7 3 1
Output: 1 2 4 8 5 3 6 7
Explanation: For the given postorder and
inorder traversal of tree the  resultant
binary tree will be
           1
       /      \
     2         3
   /  \      /  \
  4    5    6    7
   \
     8         
                 
        */
        //  
      int search(int *in,int x,int n){
        for(int i=0;i<n;i++){
            if(in[i]==x){
                return i;
            }
        }
        return -1;
    }
Node *buildTree(int in[], int pre[], int n) {
  
      if(n==0)return NULL;
        Node*root=new Node(pre[n-1]);
    int i=search(in,pre[n-1],n);
  
      root->left=buildTree(in,pre,i);
      root->right=buildTree(in+i+1,pre+i,n-i-1);
  
  
    return root;
    // Your code here
}
               /*
             Largest BST 
Medium Accuracy: 30.88% Submissions: 48314 Points: 4
Given a binary tree. Find the size of its largest subtree that is a Binary Search Tree.
Note: Here Size is equal to the number of nodes in the subtree.

Example 1:

Input:
        1
      /   \
     4     4
   /   \
  6     8
Output: 1
Explanation: There's no sub-tree with size
greater than 1 which forms a BST. All the
leaf Nodes are the BSTs with size equal
to 1.  
               
               
               */
  int count(Node *root)
   {
       if(!root)
       return 0;
    return 1 + count(root->left) + count(root->right);
   }
    bool check(Node *root,int min,int max)
   {
       if(!root)
       return true;
       bool p = check(root->left,min,root->data);
       bool k = check(root->right,root->data,max);
       int h = root->data;
       if(h>min && h<max && p && k )
       return true;
       return false;
   }
  void fun(Node*root,int &ans){
      if(root==NULL)return;
      int num=count(root);
      if(check(root,INT_MIN,INT_MAX)){
          ans=max(ans,num);
      }
         fun(root->left,ans);
       fun(root->right,ans);
  }  
    int largestBst(Node *root)
    {
        int ans=1;
        fun(root,ans);
        return ans;
/*
Merge two BST 's 
Hard Accuracy: 50.0% Submissions: 26384 Points: 8
Given two BSTs, return elements of both BSTs in sorted form.


Example 1:

Input:
BST1:
       5
     /   \
    3     6
   / \
  2   4  
BST2:
        2
      /   \
     1     3
            \
             7
            /
           6
Output: 1 2 2 3 3 4 5 6 6 7
Explanation: 
After merging and sorting the
two BST we get 1 2 2 3 3 4 5 6 6 7.
    	//Your code here
    }               

*/
   void inorder(Node*root,vector<int>&v){
        
    if(root==NULL)return;
    inorder(root->left,v);
    v.push_back(root->data);
    inorder(root->right,v);
}
vector<int>merge(vector<int>v,vector<int>s){
    vector<int>ve;
    int i=0;
    int j=0;
   int  n=v.size()>s.size()?s.size():v.size();
   
while(i<n&& j<n){
    if(v[i]<s[j]){
        ve.push_back(v[i]);
        i++;
    }
    else{
         ve.push_back(s[j]);
        j++;
    }
    
}
while(i<v.size()){
       ve.push_back(v[i]);
       i++;
}
while(j<s.size()){
       ve.push_back(s[j]);
       j++;
}
return ve;
}
    vector<int> merge(Node *root1, Node *root2)
    {
        vector<int>v;
    vector<int>s;
    inorder(root1,v);
    inorder(root2,s);
    vector<int>res=merge(v,s);
    return res;
    }
      /*
      Maximum Path Sum between 2 Leaf Nodes 
Hard Accuracy: 49.92% Submissions: 100k+ Points: 8
Given a binary tree in which each node element contains a number. Find the maximum possible path sum from one leaf node to another leaf node.

Note: Here Leaf node is a node which is connected to exactly one different node.


Example 1:

Input:      
           3                               
         /    \                          
       4       5                     
      /  \      
    -10   4                          
Output: 16
Explanation:
Maximum Sum lies between leaf node 4 and 5.
4 + 4 + 3 + 5 = 16.
Example 2:

Input:    
            -15                               
         /      \                          
        5         6                      
      /  \       / \
    -8    1     3   9
   /  \              \
  2   -3              0
                     / \
                    4  -1
                       /
                     10  
Output:  27
Explanation:
The maximum possible sum from one leaf node 
to another is (3 + 6 + 9 + 0 + -1 + 10 = 27)
      */         
 int leafsum(Node*root,int &maxi){
    if(root==NULL)return 0;
    if(root->left==NULL && root->right==NULL){
        return root->data;
    }
    int l=leafsum(root->left,maxi);
    int r=leafsum(root->right,maxi);
    int sum=l+r+root->data;
    if(root->left && root->right){
          maxi=max(maxi,sum);
          return max(l,r)+root->data;
    }
  if(root->left==NULL){
      return root->data+r;
  }
   if(root->right==NULL){
      return root->data+l;
  }
}
    int maxPathSum(Node* root)
    {
        
        int maxi=INT_MIN;
        int ans=leafsum(root,maxi);
         if(maxi==INT_MIN)
         return ans;
         return maxi;       
    }      
      /*
      Burning Tree 
Hard Accuracy: 41.61% Submissions: 6434 Points: 8
Given a binary tree and a node called target. Find the minimum time required to burn the complete binary tree if the target is set on fire. It is known that in 1 second all nodes connected to a given node get burned. That is its left child, right child, and parent.


Example 1:

Input:      
          1
        /   \
      2      3
    /  \      \
   4    5      6
       / \      \
      7   8      9
                   \
                   10
Target Node = 8
Output: 7
Explanation: If leaf with the value 
8 is set on fire. 
After 1 sec: 5 is set on fire.
After 2 sec: 2, 7 are set to fire.
After 3 sec: 4, 1 are set to fire.
After 4 sec: 3 is set to fire.
After 5 sec: 6 is set to fire.
After 6 sec: 9 is set to fire.
After 7 sec: 10 is set to fire.
It takes 7s to burn the complete tree.
      
      */   
   int fun(Node*target,map<Node*,Node*>&mp){
       queue<Node*>q;
       q.push(target);
       map<Node*,int>m;
       m[target]=1;
     
       int maxi=0;
       while(!q.empty()){
           int size=q.size();
             int fl=0;
           while(size--){
               Node*temp=q.front();
               q.pop();
               if(temp->left && !m[temp->left]){
                  fl=1;
                  m[temp->left]=1;
                  q.push(temp->left);
               }
                if(temp->right && !m[temp->right]){
                  fl=1;
                  m[temp->right]=1;
                  q.push(temp->right);
               }
               if(mp[temp] && !m[mp[temp]]){
                   fl=1;
                   m[mp[temp]]=1;
                   q.push(mp[temp]);
               }
           }
           if(fl==1)maxi++;
       }
       return maxi;
   }
  Node*solve(Node*root,map<Node*,Node*>&mp,int target){
      queue<Node*>q;
      q.push(root);
      Node*res;
      while(!q.empty()){
          Node*temp=q.front();
          q.pop();
          if(temp->data==target){
              res=temp;
          }
          if(temp->left){
              mp[temp->left]=temp;
              q.push(temp->left);
          }
           if(temp->right){
              mp[temp->right]=temp;
              q.push(temp->right);
          }
      }
      return res;
  }
    int minTime(Node* root, int target) 
    {
        
        map<Node*,Node*>mp;
     Node*targ=solve(root,mp,target); 
     return fun(targ,mp);
    }    
      /*
      Nodes at given distance in binary tree 
Hard Accuracy: 45.23% Submissions: 23906 Points: 8
Given a binary tree, a target node in the binary tree, and an integer value k, find all the nodes that are at distance k from the given target node.
No parent pointers are available.

Example 1:

Input:      
          20
        /    \
      8       22 
    /   \
   4    12 
       /   \
      10    14
Target Node = 8
K = 2
Output: 10 14 22
Explanation: The three nodes at distance 2
from node 8 are 10, 14, 22.
      */         
vector<int>fun(Node*target,map<Node*,Node*>&mp,int k){
       queue<Node*>q;
       vector<int>v;
       q.push(target);
       map<Node*,int>m;
       m[target]=1;
      int fl=0;
       int maxi=0;
       while(!q.empty()){
           int size=q.size();
            if(fl++==k)break;
           while(size--){
               Node*temp=q.front();
               q.pop();
               if(temp->left && !m[temp->left]){
                  
                  m[temp->left]=1;
                  q.push(temp->left);
               }
                if(temp->right && !m[temp->right]){
                 
                  m[temp->right]=1;
                  q.push(temp->right);
               }
               if(mp[temp] && !m[mp[temp]]){
                  
                   m[mp[temp]]=1;
                   q.push(mp[temp]);
               }
           }
          
       }
     while(!q.empty()){
         Node*curr=q.front();
          q.pop();
         v.push_back(curr->data);
        
     }
     sort(v.begin(),v.end());
     return v;
   }
 Node*solve(Node*root,map<Node*,Node*>&mp,int target){
      queue<Node*>q;
      q.push(root);
      Node*res;
      while(!q.empty()){
          Node*temp=q.front();
          q.pop();
          if(temp->data==target){
              res=temp;
          }
          if(temp->left){
              mp[temp->left]=temp;
              q.push(temp->left);
          }
           if(temp->right){
              mp[temp->right]=temp;
              q.push(temp->right);
          }
      }
      return res;
  }
    vector <int> KDistanceNodes(Node* root, int target , int k)
    {
    map<Node*,Node*>mp;
        Node*targ=solve(root,mp,target);
        vector<int>v=fun(targ,mp,k);
        return v;
        // return the sorted vector of all nodes at k dist
    }
/*
Number of Turns in Binary Tree 
Hard Accuracy: 36.4% Submissions: 8027 Points: 8
Given a binary tree and data value of two of its nodes. Find the number of turns needed to reach from one node to another in the given binary tree.

Example 1:

Input:      
Tree = 
           1
        /    \
       2       3
     /  \     /  \
    4    5   6    7                        
   /        / \                        
  8        9   10   
first node = 5
second node = 10
Output: 4
Explanation: 
Turns will be at 2, 1, 3, 6.

*/
   struct Node* lca(struct Node* root, 
                        int n1, int n2)
{
    
    if (root == NULL)
        return NULL;

   
    if (root->data == n1 || root->data == n2)
        return root;

    // Look for keys in left and right subtrees
    Node* left_lca =lca(root->left, n1, n2);
    Node* right_lca = lca(root->right, n1, n2);

  
    if (left_lca && right_lca)
        return root;

    return (left_lca != NULL) ? left_lca : 
                                right_lca;
}
   //bool CountTurn(Node*root,)
   int fun(Node*root,int a,int b,int c,char d){
       if(!root)
           return -1;
           if(root->data==a || root->data==b)
           return c;
           int left;
           if(d=='l'){
               left=fun(root->left,a,b,c,d);
               return left!=-1?left:fun(root->right,a,b,c+1,'r');
           }
           else{
               left=fun(root->left,a,b,c+1,'l');
               return left!=-1?left:fun(root->right,a,b,c,d);
           }
       }
   
    // function should return the number of turns required to go from first node to second node 
    int NumberOFTurns(struct Node* root, int first, int second)
    {
      // Your code goes here
      Node *LCA=lca(root,first,second);
  
    int turns;
    if (LCA->data==first){
        turns=fun(LCA->left,second,second,0,'l');
        turns=(turns!=-1)?turns:fun(LCA->right,second,second,0,'r');
        return (turns==0)?-1:turns;
    }
    else if(LCA->data==second){
        turns=fun(LCA->left,first,first,0,'l');
        turns=(turns!=-1)?turns:fun(LCA->right,first,first,0,'r');
        return (turns==0)?-1:turns;
    }
    int a=fun(LCA->left,first,second,0,'l');
    int b=fun(LCA->right,first,second,0,'r');
    return a+b+1;
    }            
/*
Distribute candies in a binary tree 
Hard Accuracy: 62.78% Submissions: 491 Points: 8
Given a binary tree with N nodes, in which each node value represents number of candies present at that node. In one move, one may choose two adjacent nodes and move one candy from one node to another (the move may be from parent to child, or from child to parent.) 
The task is to find the number of moves required such that every node have exactly one candy.

Example 1:

Input :      3
           /   \
          0     0 
Output : 2
Explanation: 
From the root of the tree, we move one 
candy to its left child, and one candy to
its right child

*/
   
                   int helper(Node* root, int& ans){
        if(root==NULL) return 0;
        int l = helper(root->left, ans);
        int r = helper(root->right, ans);
        ans += abs(l) + abs(r);
        return root->key + l + r - 1;
    }
    
    int distributeCandy(Node* root)
    {
        int ans = 0;
        helper(root, ans);
        return ans;
        //code here
    }
/*
Given a Binary Search Tree (BST) and a range [min, max], remove all keys which are outside the given range. The modified tree should also be BST.

Example 1:

Input:
Range = [-10, 13]
Output:
-8 6 7 13
Explanation:
Nodes with values -13, 14 and 15 are 
outside the given range and hence are 
removed from the BST.
*/
    Node* removekeys(Node* root, int min, int max) {
        if(root==NULL){
            return root;
        }
        root->left=removekeys(root->left,min,max);
        root->right=removekeys(root->right,min,max);
        if(root->data<min){
            return root->right;
        }
        if(root->data>min){
            return root->left;
        }
        return root;
    }
               /*
               Sum of k smallest elements in BST 
Easy Accuracy: 64.4% Submissions: 1637 Points: 2
Given a Binary Search Tree. Find sum of all elements smaller than and equal to Kth smallest element.

Example 1:

Input: 
          20
        /    \
       8     22
     /    \
    4     12
         /    \
        10    14   , K=3

Output: 22
Explanation:
Sum of 3 smallest elements are: 
4 + 8 + 10 = 22
               
               
               */
 void inorder(Node*root,int &k,int &sum){

    if(root==NULL)return ;
    inorder(root->left,k,sum);
    if(k!=0){
        sum+=root->data;
        k--;
    }
    inorder(root->right,k,sum);
}
int sum(Node* root, int k) 
{ 
  int sum=0;
  inorder(root,k,sum);
  return sum;
    // Your code here
     
}    
   /*
   Insert a node in a BST 
Easy Accuracy: 32.7% Submissions: 68855 Points: 2
Given a BST and a key K. If K is not present in the BST, Insert a new Node with a value equal to K into the BST. 
Note: If K is already present in the BST, don't modify the BST.


Example 1:

Input:
     2
   /   \
  1     3
K = 4
Output: 1 2 3 4
Explanation: After inserting the node 4
Inorder traversal will be 1 2 3 4.
   */
    Node* insert(Node* root, int key) {
       if(root==NULL){
        return new Node(key);
    }
    if(root->data>key){
       root->left= insert(root->left,key);
    }
    else if(key>root->data){
     root->right=insert(root->right,key);
    }
    return root;
    // Your code here
}   
               /*
               Delete a node from BST 
Medium Accuracy: 34.76% Submissions: 57365 Points: 4
Given a Binary Search Tree and a node value X. Delete the node with the given value X from the BST. If no node with value x exists, then do not make any change. 

Example 1:

Input:
      2
    /   \
   1     3
X = 12
Output: 1 2 3
Explanation: In the given input there
is no node with value 12 , so the tree
will remain same.
               
               */
               
 /*
 Find a pair with given target in BST 
Medium Accuracy: 47.17% Submissions: 30267 Points: 4
Given a Binary Search Tree and a target sum. Check whether there's a pair of Nodes in the BST with value summing up to the target sum. 

Example 1:

Input:
        2
      /   \
     1     3
sum = 5
Output: 1 
Explanation: 
Nodes with value 2 and 3 sum up to 5.
 */             
  void solve(Node*root,int target,unordered_set<int>&s,int &ans){
        if(root==NULL){
            return ;
        }
        solve(root->left,target,s,ans);
        s.insert(root->data);
        if(s.find(target-root->data)!=s.end()){
            ans=1;
            return;
        }
        solve(root->right,target,s,ans);
    
    }
    int isPairPresent(struct Node *root, int target)
    {
    unordered_set<int>s;
    int ans=0;
    solve(root,target,s,ans);
    return ans;
    //add code here.
    }    
   /*
   Fixing Two nodes of a BST 
Hard Accuracy: 50.0% Submissions: 27511 Points: 8
You are given the root of a binary search tree(BST), where exactly two nodes were swapped by mistake. Fix (or correct) the BST by swapping them back. 
Do not change the structure of the tree.
Note: It is guaranteed that the given input will form BST, except for 2 nodes that will be wrong. All changes must be reflected in the original linked list.
 
Example 1:
Input:
       10
     /    \
    5      8
   / \
  2   20
Output: 1
   */ 
   Node*&prev,Node*&first,Node*&last){
   if(root==NULL)
   return;
   compute(root->left,prev,first,last);
   if(prev!=NULL&&prev->data>root->data){
       if(first==NULL){
           first=prev;
       }
       last=root;
   }
   prev = root;
   compute(root->right,prev,first,last);
}
class Solution {
 public:
   void correctBST( struct Node* root )
   {
       // add code here.
       Node *prev=NULL,*first=NULL,*last=NULL;
       compute(root,prev,first,last);
       swap(first->data,last->data);
   }
};