【leetcode刷题】T153-二叉搜索树中的搜索

例如，
给定二叉搜索树:

        4
       / \
      2   7
     / \
    1   3

和值: 2
你应该返回如下子树:

      2     
     / \   
    1   3
在上述示例中，如果要找的值是 5，但因为没有节点值为 5，我们应该返回 NULL。

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def searchBST(self, root, val):
        """
        :type root: TreeNode
        :type val: int
        :rtype: TreeNode
        """
        node = root
        while node:
            if node.val == val:
                return node
            if node.val > val:
                node = node.left
            else:
                node = node.right
        return None

/**
 * 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* searchBST(TreeNode* root, int val) {
        TreeNode* node = root;
        while(node){
            if(node->val == val)
                return node;
            if(node->val > val)
                node = node->left;
            else
                node = node->right;
        }
        return NULL;
    }
};