Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
confused what "{1,#,2,3}"
means? >
read more on how binary tree is serialized on OJ.
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public static void copyBST(TreeNode root,List<Integer> list){ if(root==null) return; copyBST(root.left,list); list.add(root.val); copyBST(root.right,list); } public boolean isValidBST(TreeNode root) { List<Integer> list=new ArrayList<Integer>(); copyBST(root,list); for(int i=1;i<list.size();i++){ if(list.get(i)<=list.get(i-1)) return false; } return true; } }思路:中序遍历二叉搜索数,将每个节点的值存在数组中,则数组是已排序的。
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原文地址:http://blog.csdn.net/dutsoft/article/details/37727479