# Lowest Common Ancestor Of A Binary Tree Problem

## Description

LeetCode Problem 236.

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:

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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:

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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:

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

Constraints:

• The number of nodes in the tree is in the range [2, 10^5].
• -10^9 <= Node.val <= 10^9
• All Node.val are unique.
• p != q
• p and q will exist in the tree.

## Sample C++ Code

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/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool dfs(TreeNode* curr, TreeNode* p, TreeNode* q, TreeNode*& ans) {
if (curr == NULL)
return false;

int left = dfs(curr->left, p, q, ans);
int right = dfs(curr->right, p, q, ans);
int mid = ((curr == p) || (curr == q));

if (mid + left + right >= 2)
ans = curr;

return (mid + left + right) > 0;

}

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
TreeNode* anc;
dfs(root, p, q, anc);
return anc;
}
};
``````