> For the complete documentation index, see [llms.txt](https://zhenchaogan.gitbook.io/leetcode-solution/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://zhenchaogan.gitbook.io/leetcode-solution/leetcode-236-lowest-common-ancestor-of-a-binary-tree.md).

# LeetCode 236. Lowest Common Ancestor of a Binary Tree

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**).”

**Example 1:**![](https://assets.leetcode.com/uploads/2018/12/14/binarytree.png)

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

```
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]`.
* `-10^9 <= Node.val <= 10^9`
* All `Node.val` are **unique**.
* `p != q`
* `p` and `q` will exist in the tree.

## Solution

[English Version in Youtube](https://youtu.be/XhLMEiZWEZ4)

[中文版解答Youtube Link](https://youtu.be/T-PGz0jnAHA)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV14B4y1P7PF/)

```
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (root == nullptr) {
            return nullptr;
        }
    
        if (root == p || root == q) {
            return root;
        }
    
        TreeNode* l = lowestCommonAncestor(root->left, p, q);
        TreeNode* r = lowestCommonAncestor(root->right, p, q);
    
        if (l && r) {
            return root;
        }
    
        return l ? l : r;
    }
};
```


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