98 Validate Binary Search Tree 验证二叉搜索树
Description:
Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and right subtrees must also be binary search trees.
Example:
Example 1:
2
/ \
1 3
Input: [2,1,3]
Output: true
Example 2:
5
/ \
1 4
/ \
3 6
Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.
题目描述:
给定一个二叉树，判断其是否是一个有效的二叉搜索树。
假设一个二叉搜索树具有如下特征：
节点的左子树只包含小于当前节点的数。
节点的右子树只包含大于当前节点的数。
所有左子树和右子树自身必须也是二叉搜索树。
示例 :
示例 1:
输入:
2
/ \
1 3
输出: true
示例 2:
输入:
5
/ \
1 4
/ \
3 6
输出: false
解释: 输入为: [5,1,4,null,null,3,6]。
根节点的值为 5 ，但是其右子节点值为 4 。
思路:
注意这里的 BST树是对应全局递归定义的, 不能只比较当前节点的左右孩子的大小
- 递归法
给定一个范围, 左右子树都必须在该范围中 - 迭代法
相当于进行中序遍历, 要求中序遍历的结果是递增的
时间复杂度O(n), 空间复杂度O(n)
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:
bool isValidBST(TreeNode* root)
{
stack s;
long last = (long)INT_MIN - 1;
while (s.size() or root)
{
while (root)
{
s.push(root);
root = root -> left;
}
root = s.top();
s.pop();
if (root -> val <= last) return false;
last = root -> val;
root = root -> right;
}
return true;
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isValidBST(TreeNode root) {
return isValidBST(root, null, null);
}
private boolean isValidBST(TreeNode root, Integer lower, Integer upper) {
return (root == null) || ((lower == null || root.val > lower) && (upper == null || root.val < upper) && isValidBST(root.right, root.val, upper) && isValidBST(root.left, lower, root.val));
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def isValidBST(self, root: TreeNode) -> bool:
def helper(root: TreeNode, lower: int, upper: int) -> bool:
if not root:
return True
return (lower < root.val < upper) and helper(root.left, lower, root.val) and helper(root.right, root.val, upper)
return helper(root, -(1 << 33), 1 << 33)