思路
首先搞清楚二叉搜索树的机制,左儿子的值都小于当前节点,右儿子的值都大于当前节点
然后就可以快速写出程序了,用递归很好实现
1. 相等直接return
2. 小于当前节点,直接去树节点的左子树寻找
3. 大于当前节点,去树节点的右子树寻找
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def searchBST(self, root: TreeNode, val: int) -> TreeNode:
if root is None:
return None
if val == root.val:
return root
elif val > root.val:
return self.searchBST(root.right, val)
else:
return self.searchBST(root.left, val)
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func searchBST(root *TreeNode, val int) *TreeNode {
if root == nil {
return nil
}
if val < root.Val {
return searchBST(root.Left, val)
}
if val > root.Val {
return searchBST(root.Right, val)
}
return root
}