98. Validate Binary Search Tree

题目：

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.

链接：http://leetcode.com/problems/validate-binary-search-tree/

5/9/2017

算法班

自己之前没做出来，参考的别人答案。

思路：用in-order traversal判断是否是BST

public class Solution {
    TreeNode lastNode;
    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;
        if (!isValidBST(root.left)) return false;
        if (lastNode != null && lastNode.val >= root.val) return false;
        lastNode = root;
        if (!isValidBST(root.right)) return false;
        return true;
    }
}

iterative的写法，不需要全局变量

三种tree traversal，敲黑板

public class Solution {
    public boolean isValidBST(TreeNode root) {
        // Just use the inOrder traversal to solve the problem.
        if (root == null) {
            return true;
        }
        
        Stack<TreeNode> s = new Stack<TreeNode>();
        TreeNode cur = root;
        TreeNode prev = null;
        
        while (cur != null || !s.empty()) {
            while (cur != null) {
                s.push(cur);
                cur = cur.left;
            }
            cur = s.pop();
            if (prev != null && prev.val >= cur.val) return false;
            prev = cur;
            cur = cur.right;
        }
        return true;
    }
}

有些其他人的算法是传入long.MAX_VALUE, long.MIN_VALUE来判断的，我认为算是cheat吧，至少需要知道不使用这种trick的方法。

更多讨论：

https://discuss.leetcode.com/category/106/validate-binary-search-tree