98. Validate Binary Search Tree

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.

思路：中序排列为排序好的数组（数组元素一次增大）；数组中不能有重复元素。

代码如下：

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public boolean isValidBST(TreeNode root) {

         if(root==null||(root.left==null&&root.right==null))
         return true;
         ArrayList<Integer> list=ParentAndSon(root);
           int[] nums=new int[list.size()];
           int[] nums1=new int[list.size()];
           for(int i=0;i<list.size();i++)
           {
               nums[i]=list.get(i);
               nums1[i]=list.get(i);
           }
           Arrays.sort(nums);

           if(nums[0]!=nums1[0])
           return false;
           for(int i=1;i<nums.length;i++)
           {
            if(nums1[i]!=nums[i]||nums[i]==nums[i-1])
            return false;

           }
      return true;
     }
     public ArrayList<Integer> ParentAndSon(TreeNode root){
         ArrayList<Integer> list=new ArrayList<>();
         if(root.left!=null)
         list.addAll(ParentAndSon(root.left));

         list.add(root.val);

         if(root.right!=null)
         list.addAll(ParentAndSon(root.right));
         return list;
     }
 }