Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

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 binary search tree:  root = [6,2,8,0,4,7,9,null,null,3,5]

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

题目大意：

　　给定一个二叉树，和二叉树中的两个节点，判断这两个节点的公共祖先。

理  解：

　　两个节点的两种位置，当前根节点的左右子树中或者两个节点是父子关系。

代 码 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== NULL)
            return NULL;
        if(p->val == root->val || q->val == root->val)
            return root;
        TreeNode *left,*right;
        left  = lowestCommonAncestor(root->left,p,q);
        right = lowestCommonAncestor(root->right,p,q);
        if(left!=NULL && right!=NULL)
            return root;
        else if(left==NULL)
            return right;
        else if(right==NULL)
            return left;
        else
            return NULL;
    }
};

运行结果：

　　执行用时 :76 ms, 在所有C++提交中击败了35.68%的用户

　　内存消耗 :26 MB, 在所有C++提交中击败了5.28%的用户