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:

1

 root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1

Output:

1

 3

Explanation:

The LCA of nodes 5 and 1 is 3.

Example 2:

Input:

1

 root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4

Output:

1

 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)  
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/**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;
    }
};