【leetcode】Validate Binary Search Tree（middle）-优快云博客

本文介绍了一种通过中序遍历来验证给定二叉树是否为有效二叉搜索树的方法。同时提供了递归版的实现方式，强调了左子树值范围在最小值到根值之间，右子树值范围在根值到最大值之间的特性。

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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.

思路：中序遍历。当前值要比之前的小。

bool isValidBST(TreeNode* root) {
        TreeNode * pPre = NULL;
        TreeNode * pCur = root;
        vector<TreeNode *> v;

        while(!v.empty() || NULL != pCur)
        {
            if(NULL != pCur)
            {
                v.push_back(pCur);
                pCur = pCur->left;
            }
            else
            {
                if(pPre != NULL && v.back()->val <= pPre->val)
                    return false;
                pPre = v.back();    
                v.pop_back();
                pCur = pPre->right;
            }
        }
        return true;
    }

大神递归版：注意，每次左子树的值范围在最小值和根值之间，右子树的范围在根植和最大值之间。

public class Solution {
    public boolean isValidBST(TreeNode root) {
        return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }

    public boolean isValidBST(TreeNode root, long minVal, long maxVal) {
        if (root == null) return true;
        if (root.val >= maxVal || root.val <= minVal) return false;
        return isValidBST(root.left, minVal, root.val) && isValidBST(root.right, root.val, maxVal);
    }
}