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53_lowest-common-ancestor-of-a-binary-search-tree.cpp
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53_lowest-common-ancestor-of-a-binary-search-tree.cpp
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// DATE: 07-Aug-2023
/* PROGRAM: 53_Tree - Lowest Common Ancestor of BST
https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in
the BST.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two
nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to
be a descendant of itself).”
Example 1:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.
Example 2:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to
the LCA definition.
Example 3:
Input: root = [2,1], p = 2, q = 1
Output: 2
*/
// @ankitsamaddar @Aug_2023
#include <iostream>
#include <vector>
using namespace std;
struct TreeNode {
int val;
TreeNode* left;
TreeNode* right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode* left, TreeNode* right) : val(x), left(left), right(right) {}
};
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
TreeNode* cur = root;
while (cur) {
if ((p->val > cur->val) and (q->val > cur->val)) {
cur = cur->right;
} else if ((p->val < cur->val) and (q->val < cur->val)) {
cur = cur->left;
} else {
break;
}
}
return cur;
}
};
int main() {
TreeNode* root = new TreeNode(6);
root->left = new TreeNode(2);
root->right = new TreeNode(8);
// left sub-tree
root->left->left = new TreeNode(0);
root->left->right = new TreeNode(4);
root->left->right->left = new TreeNode(3);
root->left->right->right = new TreeNode(5);
// right sub-tree
root->right->left = new TreeNode(7);
root->right->right = new TreeNode(9);
// p and q
TreeNode* p = new TreeNode(2);
TreeNode* q = new TreeNode(8);
Solution sol;
TreeNode* res = sol.lowestCommonAncestor(root, p, q);
cout << res->val; // Output: 6
return 0;
}