【重点】【二叉树】98.验证二叉搜索树

题目

Python

参考：https://leetcode.cn/problems/validate-binary-search-tree/solutions/2020306/qian-xu-zhong-xu-hou-xu-san-chong-fang-f-yxvh/?envType=study-plan-v2&envId=top-100-liked
前序遍历！！！

# 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 isValidBST(self, root: Optional[TreeNode]) -> bool:
        return self.judge(root, -inf, inf)

    def judge(self, root, leftVal, rightVal):
        if not root:
            return True
        x = root.val
        return (leftVal < x < rightVal
                and self.judge(root.left, leftVal, x)
                and self.judge(root.right, x, rightVal))

Java

class Solution {
    public boolean isValidBST(TreeNode root) {
        if (root == null || (root.left == null && root.right == null)) {
            return true;
        }
        TreeNode maxLeftNode = root.left;   // 左子树最大值
        TreeNode minRightNode = root.right; // 右子树最小值
        while (maxLeftNode != null && maxLeftNode.right != null) {
            maxLeftNode = maxLeftNode.right;
        }
        while (minRightNode != null && minRightNode.left != null) {
            minRightNode = minRightNode.left;
        }

        if (root.left == null) {
            return (root.val < minRightNode.val) && isValidBST(root.right);
        } else if (root.right == null) {
            return (maxLeftNode.val < root.val) && isValidBST(root.left);
        }

        return (maxLeftNode.val < root.val) 
                && (root.val < minRightNode.val)
                && isValidBST(root.left) 
                && isValidBST(root.right);
    }
}