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# Lowest Common Ancestor of a Binary Tree LeetCode Solution

## Problem – 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.

### Lowest Common Ancestor of a Binary Tree LeetCode Solution in C++

``````TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if (!root || root == p || root == q) return root;
TreeNode* left = lowestCommonAncestor(root->left, p, q);
TreeNode* right = lowestCommonAncestor(root->right, p, q);
return !left ? right : !right ? left : root;
}
``````

### Lowest Common Ancestor of a Binary Tree LeetCode Solution in Python

``````def lowestCommonAncestor(self, root, p, q):
if root in (None, p, q): return root
left, right = (self.lowestCommonAncestor(kid, p, q)
for kid in (root.left, root.right))
return root if left and right else left or right
``````

### Lowest Common Ancestor of a Binary Tree LeetCode Solution in Java

``````public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null || root == p || root == q) return root;
TreeNode left = lowestCommonAncestor(root.left, p, q);
TreeNode right = lowestCommonAncestor(root.right, p, q);
return left == null ? right : right == null ? left : root;
}
``````
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