# LeetCode-Lowest Common Ancestor of a Binary Tree  Solution
Date: 2020-07-28


### **题目：**

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 = \[3,5,1,6,2,0,8,null,null,7,4\]

![](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.

### **实现：**

- 递归方式，从根节点开始查找 p和q，如果查找到直接返回root，不是从左右子树分配查找， 如果左子树为空，那么找右子树， 如果右子树为空找左子树。 
-  时间复杂度 O(N)  

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