【Leetcode】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.

confused what 

"{1,#,2,3}"

 means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

1
  / \
 2   3
    /
   4
    \
     5
      

The above binary tree is serialized as 

"{1,2,3,#,#,4,#,#,5}"

.

Java:

http://blog.csdn.net/linhuanmars/article/details/23810735

其實感覺第二種方法更好了解，第一種不太懂，有懂的朋友幫忙解釋下：

第一種：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isValidBST(TreeNode root) {
        ArrayList<Integer> pre=new ArrayList<Integer>();
        pre.add(null);
        return help(root,pre);
    }
    private boolean help(TreeNode root,ArrayList<Integer> pre)
    {
        if(root==null) return true;
       boolean left=help(root.left,pre);
       if(pre.get(0)!=null&&root.val<=pre.get(0)) return false;
       pre.set(0,root.val);
        return left&&help(root.right,pre);
    }
}
           

第二種：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isValidBST(TreeNode root) {
        return help(root,Integer.MIN_VALUE, Integer.MAX_VALUE);
    }
    private boolean help(TreeNode root,int min,int max)
    {
        if(root==null) return true;
        if(root.val<=min||root.val>=max) return false;
        return help(root.left,min,root.val)&&help(root.right, root.val, max);
    }
}