---
# System prepended metadata

title: 236. Lowest Common Ancestor of a Binary Tree
tags: [' bt', ' medium', tree]

---

---
link: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/
tags: tree, bt, medium
---

# 236. Lowest Common Ancestor of a Binary Tree

## Question
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](https://en.wikipedia.org/wiki/Lowest_common_ancestor): “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**).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

![img](https://assets.leetcode.com/uploads/2018/12/14/binarytree.png)

**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.
```

 

**Note:**

- All of the nodes' values will be unique.
- p and q are different and both values will exist in the binary tree.

## Solution: Python
```python=
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        self.lca = None
        self.helper(root, p, q)
        return self.lca
        
    def helper(self, node, p, q):
        if node is None:
            return False, False
        
        l_found_p, l_found_q = self.helper(node.left, p, q)
        r_found_p, r_found_q = self.helper(node.right, p, q)
        n_found_p, n_found_q = node == p, node == q
        
        found_p = (l_found_p or r_found_p or n_found_p)
        found_q = (l_found_q or r_found_q or n_found_q)
                
        if found_p and found_q and self.lca == None:
            self.lca = node
        
        return found_p, found_q
        
```

## Solution: Java
```java=        
```