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

 1 public class Solution {
 2     TreeNode lastNode;
 3     public boolean isValidBST(TreeNode root) {
 4         if (root == null) return true;
 5         if (!isValidBST(root.left)) return false;
 6         if (lastNode != null && lastNode.val >= root.val) return false;
 7         lastNode = root;
 8         if (!isValidBST(root.right)) return false;
 9         return true;
10     }
11 }

iterative的写法，不需要全局变量

三种tree traversal，敲黑板

 1 public class Solution {
 2     public boolean isValidBST(TreeNode root) {
 3         // Just use the inOrder traversal to solve the problem.
 4         if (root == null) {
 5             return true;
 6         }
 7         
 8         Stack<TreeNode> s = new Stack<TreeNode>();
 9         TreeNode cur = root;
10         TreeNode prev = null;
11         
12         while (cur != null || !s.empty()) {
13             while (cur != null) {
14                 s.push(cur);
15                 cur = cur.left;
16             }
17             cur = s.pop();
18             if (prev != null && prev.val >= cur.val) return false;
19             prev = cur;
20             cur = cur.right;
21         }
22         return true;
23     }
24 }

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。