leetcode oj java 98. Validate Binary Search Tree

本文主要是介绍leetcode oj java 98. Validate Binary Search Tree，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

一、问题描述：

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.

二、解决思路：

     二分搜索树的中序遍历应该是一个有序的数组，把树中序遍历判断数组是否有序

三、代码：

package T12;import java.util.ArrayList;
import java.util.List;/*** @author 作者 : xcy* @version 创建时间：2016年12月30日 下午1:58:43*          类说明*/
public class t98 {public static void main(String[] args) {// TODO Auto-generated method stubTreeNode root = new TreeNode(10);TreeNode r1 = new TreeNode(6);TreeNode r2 = new TreeNode(14);TreeNode r3 = new TreeNode(4);TreeNode r4 = new TreeNode(8);TreeNode r5 = new TreeNode(12);TreeNode r6 = new TreeNode(16);root.left = r1;root.right = r2;r1.left = r3;r1.right = r4;r2.left = r5;r2.right = r6;System.out.println(isValidBST(root));}public static boolean isValidBST(TreeNode root) {if (root == null || (root.left == null && root.right == null)) {return true;}List<Integer> list = new ArrayList<Integer>();genList(root, list);for (int i = 0; i < list.size() - 1; i++) {if (list.get(i) > list.get(i + 1))return false;}return true;}public static void genList(TreeNode root, List<Integer> list) {if (root == null) {return;}genList(root.left, list);list.add(root.val);genList(root.right, list);}}

这篇关于leetcode oj java 98. Validate Binary Search Tree的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---