Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
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/**
* 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。Validate Binary Search Tree,布布扣,bubuko.com
原文:http://blog.csdn.net/dutsoft/article/details/37727479