[leetcode] 700. Search in a Binary Search Tree

Description

Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node’s value equals the given value. Return the subtree rooted with that node. If such node doesn’t exist, you should return NULL.

For example,

Given the tree:
        4
       / \
      2   7
     / \
    1   3

And the value to search: 2

You should return this subtree:

      2     
     / \   
    1   3

In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.

Note that an empty tree is represented by NULL, therefore you would see the expected output (serialized tree format) as [], not null.

分析

题目的意思是：在一个二叉查找树里面找出是否存在给定的值的节点。思路很直接，直接遍历就可以了。

代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def searchBST(self, root: TreeNode, val: int) -> TreeNode:
        if(root is None):
            return None
        elif(root.valval):
            return self.searchBST(root.left,val)
        else:
            return root