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,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;
    }
}

 非递归法：

/**
 * 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;
        }
        Stack<TreeNode> stack = new Stack<>();
        List<Integer> list = new ArrayList<>();
        
        TreeNode p = root;
         while(p != null || !stack.isEmpty()){
             
             if(p != null){
                 stack.push(p);
                 p = p.left;
             }
             else{
                 TreeNode t = stack.pop();
                 if(list.size() != 0 ){
                    if(t.val > list.get(list.size()-1)){
                        list.add(t.val);
                    }
                    else{
                        return false;
                    }
                }
                 else{
                     list.add(t.val);
                 }
                 p = t.right;
             }
         }
        
        return true;
        
    }
    

}

 二刷, 根据二叉树原理进行递归：

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