leetcode oj java 98. Validate Binary Search Tree

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 stub        TreeNode 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);    }}