# 236. Lowest Common Ancestor of a Binary Tree | LeetCode Solution

## 236. Lowest Common Ancestor of a Binary Tree | LeetCode Solution

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.

## The solution to the above question is -

/**
* 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:
TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *p, TreeNode *q)
{
if (root == NULL)
return root;

if (root == p || root == q)
return root;

TreeNode *left = lowestCommonAncestor(root->left, p, q);
TreeNode *right = lowestCommonAncestor(root->right, p, q);
if (left != NULL && right != NULL)
{
return root;
}
else if (left == NULL)
{
return right;
}
else
{
return left;
}
return root;
}
};