0098. Validate Binary Search Tree

https://leetcode.com/problems/validate-binary-search-tree

Description

Given the root of a binary tree, determine if it is a valid binary search tree (BST).

A valid 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:** root = [2,1,3]
**Output:** true

Example 2:

**Input:** root = [5,1,4,null,null,3,6]
**Output:** false
**Explanation:** The root node's value is 5 but its right child's value is 4.

Constraints:

• The number of nodes in the tree is in the range [1, 104].

• -231 <= Node.val <= 231 - 1

ac1: inorder traversal

• BST, inorder traversal is transcendent.

• Traversal: consider three part: left,right and itself, it's about order.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private boolean firstNode = true; // first flag to handle corner test case, like it gives you MIN_VALUE;
    private int pre = Integer.MIN_VALUE;
    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;

        // Traversal, left first
        if (!isValidBST(root.left)) return false;

        // self
        if (root.val <= pre && !firstNode) return false;
        firstNode = false;
        pre = root.val;

        // right
        if (!isValidBST(root.right)) return false;

        return true;

    }
}

ac2: iterative, same as traversal

in this case, iteration is easier to understand.

class Solution {
    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;

        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode pre = null;

        TreeNode curr = root;
        while (!stack.isEmpty() || curr != null) {
            if (curr != null) {
                stack.push(curr);
                curr = curr.left;
            } else {
                curr = stack.pop();
                if (pre != null && curr.val <= pre.val) return false;
                pre = curr;
                curr = curr.right;
            }
        }

        return true;
    }
}

First answer, so ugly, must be wrong.

/**
 * 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;
        if (root.left == null && root.right == null) {
            return true;
        }

        int left = max(root.left);
        int right = min(root.right);
        if (root.left == null && right <= root.val) return false;
        if (root.right == null && left >= root.val) return false;
        if (root.left != null && root.right != null) {
            if (left >= root.val || right <= root.val) return false;
        }

        return isValidBST(root.left) && isValidBST(root.right);
    }

    private int min(TreeNode node) {
        if (node == null) return Integer.MAX_VALUE;

        if (node.left == null && node.right == null) {
            return node.val;
        }

        int left = min(node.left);
        int right = min(node.right);

        int minChild = Math.min(left, right);
        return Math.min(node.val, minChild);    
    }

    private int max(TreeNode node) {
        if (node == null) return Integer.MIN_VALUE;

        if (node.left == null && node.right == null) {
            return node.val;
        }

        int left = max(node.left);
        int right = max(node.right);

        int maxChild = Math.max(left, right);
        return Math.max(node.val, maxChild);    
    }
}

ac3: recursive

becareful about the boudary, especially Integer.MAX_VALUE Integer.MIN_VALUE

2/18/2018

/**
 * 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) {
        // edge cases
        if (root == null) return true;

        return valid(root.left, Long.MIN_VALUE, root.val) && valid(root.right, root.val, Long.MAX_VALUE);
    }

    private boolean valid(TreeNode root, long low, long up) {
        if (root == null) return true;

        if (root.val <= low || root.val >= up) return false;

        return valid(root.left, low, root.val) && valid(root.right, root.val, up);
    }
}