判断二叉搜索树

整理思路：这个是最简单的一个思路，本人能够想到的。中序遍历二叉树，得到有序数组。

import java.util.ArrayList;
import java.util.List;

public class num_19 {

	public static class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;

		TreeNode(int x) {
			val = x;
		}
	}

	public static void main(String[] args) {
		TreeNode root1 = new TreeNode(10);
		TreeNode root2 = new TreeNode(5);
		TreeNode root3 = new TreeNode(15);
		TreeNode root4 = new TreeNode(6);
		TreeNode root5 = new TreeNode(20);
		root1.left = root2;
		root1.right = root3;
		root3.left = root4;
		root3.right = root5;
		System.out.println(isValidBST(root1));

	}

	public static boolean isValidBST(TreeNode root) {
		List<Integer> list = new ArrayList<Integer>();
		inorder(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 inorder(TreeNode node, List<Integer> list) {
		if (node == null)
			return;
		inorder(node.left, list);
		list.add(node.val);
		inorder(node.right, list);
	}

}