LeetCode 98 Validate Binary Search Tree (栈)

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本文探讨了如何判断一个二叉树是否为有效的二叉搜索树，通过中序遍历的方法，确保左子树的所有节点小于根节点，右子树的所有节点大于根节点，并且左右子树也必须是二叉搜索树。

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 1:

Input:
    2
   / \
  1   3
Output: true

Example 2:

    5
   / \
  1   4
     / \
    3   6
Output: false
Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value
             is 5 but its right child's value is 4.

题目链接：https://leetcode.com/problems/validate-binary-search-tree/

题目分析：栈中序

3ms，时间击败7%，最快是递归做法

/**
 * 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) {
        if (root == null || (root.left == null && root.right == null)) {
            return true;
        }
        TreeNode pre = null;
        Stack<TreeNode> stk = new Stack<>();
        while (root != null || !stk.empty()) {
            while (root != null) {
                stk.add(root);
                root = root.left;
            }
            if (!stk.empty()) {
                root = stk.peek();
                stk.pop();
                if (pre == null) {
                    pre = root;
                } else if (pre.val >= root.val) {
                    return false;
                }
                pre = root;
                root = root.right;
            }
        }
        return true;
    }
}