Validate Binary Search Tree LeetCode Java

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.

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Two Approach:

  Solution_1: In_Order Approach

    Idea: simply do In_Order traversal, then check
    If In_Order traversal of a binary tree is sorted,
    then the binary tree is BST. else not

public static boolean isValidBSTInOrder(TreeNode root) {
        if(root==null) return true;
        ArrayList<Integer> res=new ArrayList<Integer>();
        Stack<TreeNode> stack=new Stack<TreeNode>();
        while(!stack.empty() || root!=null){
            if(root!=null){
                stack.add(root);
                root=root.left;
            }else {
                root=stack.pop();
                res.add(root.val);
                root=root.right;
            }
        }
        //System.out.println(res);
        int num=res.get(0);
        for(int i=1;i<res.size();i++){
            if(res.get(i)<=num){
                return false;
            }
            num=res.get(i);
        }
        return true;
    }

 Solution_2:
    Key to solve: DFS Traversal with ceiling && flooring cases

return false, if there is invalid case was found

public static boolean isValidBST(TreeNode root) {
        return helper(root,Integer.MIN_VALUE,Integer.MAX_VALUE);
    }
    private static boolean helper(TreeNode root,int flooring, int ceiling){
        if(root==null) return true;

        //check for invalid case
        if(root.val>=ceiling || root.val<=flooring) return false;

        //check both flooring && ceiling case
        //case1: Left subtree nodes=> flooring: min, ceiling: root.val
        //case2: Right subtree nodes => flooring:root.val, ceiling: max
        return helper(root.left,flooring,root.val)
                && helper(root.right,root.val,ceiling);
    }