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.
Example 1:

Input:
    2
   /   1   3
Output: true
Example 2:

    5
   /   1   4
     /     3   6
Output: false
Explanation: The input is: [5,1,4,null,3,6]. The root node‘s value
             is 5 but its right child‘s value is 4.

如果对BST 中序遍历， 会产生有序数列。

/**
 * DeFinition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    //BST inorder traverse will generate a ordered sequnence
    public boolean isValidBST(TreeNode root) {
        if(root == null){
            return true;
        }
        
        List<Integer> list = new ArrayList<>();
        list.add(null);
        return helper(root,list);
    }
    
    public boolean helper(TreeNode node,List<Integer> list){
        if(node == null){
            return true;
        }
        boolean left = helper(node.left,list);
        
        if(list.get(list.size()-1) != null && node.val <= list.get(list.size()-1)){
            return false;
        }
        list.add(node.val);
        boolean right = helper(node.right,list);
        
        return left && right;
    }
}