# LeetCode-Lowest Common Ancestor of a Binary Tree Solution

### 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: “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] 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;
}
};``````