98. Validate Binary Search Tree

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:

    2
   / \
  1   3

Binary tree [2,1,3], return true.

 

Example 2:

    1
   / \
  2   3

Binary tree [1,2,3], return false.

本题采用的方法是迭代的先序遍历检验法,因为起点变成了从最左节点开始,所以可以当作没有左节点,只要比较其父节点和父节点的右子树节点大小即可,stack用来存储其父节点。代码如下:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public boolean isValidBST(TreeNode root) {
12         TreeNode node = root;
13         Stack<TreeNode> stack = new Stack<TreeNode>();
14         TreeNode pre = null;
15         while(node!=null||!stack.isEmpty()){
16             while(node!=null){
17                 stack.push(node);
18                 node = node.left;
19             }
20             node = stack.pop();
21             if(pre!=null&&pre.val>=node.val) return false;
22             pre = node;
23             node = node.right;
24         }
25         return true;
26     }
27 }

 

posted @ 2017-03-23 10:28  CodesKiller  阅读(151)  评论(0编辑  收藏  举报