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题目
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.
递归
查找树中是否存在val,那么先从跟结点开始比较，如果相等，则直接返回二叉树；如果 val 比根结点值大，说明可能在右子树中，递归查找右子树；如果val 比根结点值小，说明可能在左子树中，递归查找左子树。
python 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 not root:
return None
if root.val == val:
return root
elif root.val<val:
return self.searchBST(root.right,val)
elif root.val>val:
return self.searchBST(root.left,val)
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