D-27 submission.md

⚡ Question

Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

🍑 Solution

class Solution {
 
 HashMap<Integer,Integer> map=null;
 int post_index=0;
 
 public TreeNode buildTree(int[] inorder, int[] post) {
     map=new HashMap<>();
     post_index=post.length-1;
     
     for(int i=0;i<inorder.length;i++){
         map.put(inorder[i],i);
     }
     return construct(post,0,inorder.length-1);
 }
 
 public TreeNode construct(int[] post,int instart,int inend){
     if(instart>inend) return null;
     
     int index=map.get(post[post_index]);
             
     TreeNode root=new TreeNode(post[post_index]);
     
     post_index--;
     
     root.right=construct(post,index+1,inend);
     root.left=construct(post,instart,index-1);
      
     return root;
 }
}