LeetCode T98.Validate Binary Search Tree/验证二叉搜索树

 本题有两种思路求解,一种是利用递归,逐层验证该节点是否符合BST的要求,空间复杂度较大。第二种是利用中序遍历二叉搜索树,判断序列是否升序来判断是否为二叉搜索树。

我的题解代码如下,leetcode上运行时间16ms,内存占用18.3MB

/**
 * 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<TreeNode*> stack;
        long long inorder = (long long)INT_MIN - 1;

        while (!stack.empty() || root != nullptr) {
            while (root != nullptr) {
                stack.push(root);
                root = root -> left;
            }
            root = stack.top();
            stack.pop();
            // 中序遍历后得到的是一个升序序列
            if (root -> val <= inorder) return false;
            inorder = root -> val;
            root = root -> right;
        }
        return true;
    }
};
原文地址:https://www.cnblogs.com/runsdeep/p/12829334.html