题目链接 : https://leetcode-cn.com/problems/validate-binary-search-tree/
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:
因为二叉搜索树中序遍历是递增的,所以我们可以中序遍历判断前一数是否小于后一个数.
# 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:
res = []
def helper(root):
if not root:
return
helper(root.left)
res.append(root.val)
helper(root.right)
helper(root)
return res == sorted(res) and len(set(res)) == len(res)
思路一:迭代
我们可以通过中序遍历迭代方式94. 二叉树的中序遍历来判断.
思路二:递归
- 中序遍历递归
- 利用
max_val
和min_val
代码:
迭代
# 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:
stack = []
p = root
pre = None
while p or stack:
while p:
stack.append(p)
p = p.left
p = stack.pop()
if pre and p.val <= pre.val:
return False
pre = p
p = p.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) {
Deque<TreeNode> stack = new LinkedList<>();
TreeNode p = root;
TreeNode pre = null;
while (p != null || !stack.isEmpty()) {
while (p != null) {
stack.push(p);
p = p.left;
}
p = stack.pop();
if (pre != null && pre.val >= p.val) return false;
pre = p;
p = p.right;
}
return true;
}
}
思路二
利用递归中序遍历
# 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:
self.pre = None
def isBST(root):
if not root:
return True
if not isBST(root.left):
return False
if self.pre and self.pre.val >= root.val:
return False
self.pre = root
#print(root.val)
return isBST(root.right)
return isBST(root)
java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
TreeNode pre = null;
public boolean isValidBST(TreeNode root) {
if (root == null) return true;
if (!isValidBST(root.left)) return false;
if (pre != null && pre.val >= root.val) return false;
pre = root;
return isValidBST(root.right);
}
}
利用最大值最小值
# 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 isBST(root, min_val, max_val):
if root == None:
return True
# print(root.val)
if root.val >= max_val or root.val <= min_val:
return False
return isBST(root.left, min_val, root.val) and isBST(root.right, root.val, max_val)
return isBST(root, float("-inf"), float("inf"))
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 isBST(root, Long.MAX_VALUE, Long.MIN_VALUE);
}
private boolean isBST(TreeNode root, long maxValue, long minValue) {
if (root == null) return true;
if (root.val >= maxValue || root.val <= minValue) return false;
return isBST(root.left, root.val, minValue) && isBST(root.right, maxValue, root.val);
}
}
原文地址:https://www.cnblogs.com/powercai/p/11069820.html
时间: 2024-11-08 20:57:35