LeetCode-in-All

236. Lowest Common Ancestor of a Binary Tree

Medium

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

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 = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1

Output: 3

Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4

Output: 5

Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [1,2], p = 1, q = 2

Output: 1

Constraints:

• The number of nodes in the tree is in the range [2, 105].
• -109 <= Node.val <= 109
• All Node.val are unique.
• p != q
• p and q will exist in the tree.

Solution

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
#include <stdio.h>
#include <stdlib.h>

// Function to find the lowest common ancestor of two nodes in the binary tree
struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
    // Base case: if the root is NULL or root matches either p or q
    if (root == NULL || root == p || root == q) {
        return root;
    }

    // Recursively find the LCA in the left and right subtrees
    struct TreeNode* left = lowestCommonAncestor(root->left, p, q);
    struct TreeNode* right = lowestCommonAncestor(root->right, p, q);

    // If both left and right are non-null, the current root is the LCA
    if (left != NULL && right != NULL) {
        return root;
    }

    // Otherwise, return the non-null child (if both are NULL, return NULL)
    return left != NULL ? left : right;
}

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