Question

Formatted question description: https://leetcode.ca/all/98.html

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

Algorithm

Make use of the nature of BST, bring the maximum and minimum values of the system during initialization, and replace them with their own node values in the recursive process.

The use of long instead of int is to include the boundary conditions of int.

For below BFS solution, we can also define a new data structure MyNode

class MyNode {

    TreeNode node;
    int upper;
    int lower;
}

Then initiate as

new MyNode(node, Integer.Max, Integer.Min)

And enqueue as

queue.offer(new MyNode(node.left, lower, currentNode.val))
queue.offer(new MyNode(node.right, currentNode.val, upper))

Note: remember below in invalid, 7 should go right of 5

Code

import java.util.LinkedList;

public class Validate_Binary_Search_Tree {

	/**
	 * Definition for a binary tree node.
	 * public class TreeNode {
	 *     int val;
	 *     TreeNode left;
	 *     TreeNode right;
	 *     TreeNode(int x) { val = x; }
	 * }
	 */

    public class Solution { // dfs

        public boolean isValidBST(TreeNode root) {
            return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
        }

        public boolean isValidBST(TreeNode root, long min, long max) { // int will be overflow by input [2147483647]
            if (root == null) {
                return true;
            }
            if (root.val >= max || root.val <= min) { // @note: must be <= and >=, eg. [1,1]
                return false;
            }

            // below code, check with left/right children not necessary, since min/max updated already
            return isValidBST(root.left, min, root.val) && isValidBST(root.right, root.val, max);
        }
    }

    class Solution_iteration {

        LinkedList<TreeNode> queue = new LinkedList<>(); // store node, and its associated low/high in other two lists
        LinkedList<Integer> uppers = new LinkedList<>();
        LinkedList<Integer> lowers = new LinkedList<>();

        public void update(TreeNode root, Integer lower, Integer upper) {
            queue.add(root);
            lowers.add(lower);
            uppers.add(upper);
        }

        public boolean isValidBST(TreeNode root) {
            Integer lower = null;
            Integer upper = null;
            int val;
            update(root, lower, upper);

            while (!queue.isEmpty()) {
                root = queue.poll();
                lower = lowers.poll();
                upper = uppers.poll();

                if (root == null) continue;
                val = root.val;
                if (lower != null && val <= lower) return false;
                if (upper != null && val >= upper) return false;
                update(root.right, val, upper);
                update(root.left, lower, val);
            }
            return true;
        }
    }

}

############

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private Integer prev;

    public boolean isValidBST(TreeNode root) {
        prev = null;
        return dfs(root);
    }

    private boolean dfs(TreeNode root) {
        if (root == null) {
            return true;
        }
        if (!dfs(root.left)) {
            return false;
        }
        if (prev != null && prev >= root.val) {
            return false;
        }
        prev = root.val;
        if (!dfs(root.right)) {
            return false;
        }
        return true;
    }
}

// OJ: https://leetcode.com/problems/validate-binary-search-tree
// Time: O(N)
// Space: O(H)
class Solution {
public:
    bool isValidBST(TreeNode* root, TreeNode *left = NULL, TreeNode *right = NULL) {
        if (!root) return true;
        if ((left && root->val <= left->val) || (right && root->val >= right->val)) return false;
        return isValidBST(root->left, left, root) && isValidBST(root->right, root, right);
    }
};

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isValidBST(self, root: Optional[TreeNode]) -> bool:
        def dfs(root, mi, ma):
            if not root:
                return True
            return mi < root.val < ma and dfs(root.left, mi, root.val) and dfs(root.right, root.val, ma)
        return dfs(root, -math.inf, math.inf)


class Solution: # iterative
    def isValidBST(self, root: TreeNode) -> bool:
        # store (node, lower, upper) tuples in a single queue
        queue = [(root, None, None)]

        # another option:
        # queue = [(root, -math.inf, math.inf)]

        while queue:
            node, lower, upper = queue.pop(0)
            
            if node is None:
                continue
            
            val = node.val
            if lower is not None and val <= lower:
                return False
            if upper is not None and val >= upper:
                return False
            
            queue.append((node.right, val, upper))
            queue.append((node.left, lower, val))
        
        return True

############

class Solution:
  def isValidBST(self, root: Optional[TreeNode]) -> bool:
    def isValidBST(root: Optional[TreeNode],
                   minNode: Optional[TreeNode], maxNode: Optional[TreeNode]) -> bool:
      if not root:
        return True
      if minNode and root.val <= minNode.val:
        return False
      if maxNode and root.val >= maxNode.val:
        return False

      return isValidBST(root.left, minNode, root) and isValidBST(root.right, root, maxNode)

    return isValidBST(root, None, None)

############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
  def isValidBST(self, root):
    """
    :type root: TreeNode
    :rtype: bool
    """
    prev = -float("inf")
    stack = [(1, root)]
    while stack:
      p = stack.pop()
      if not p[1]:
        continue
      if p[0] == 0:
        if p[1].val <= prev:
          return False
        prev = p[1].val
      else:
        stack.append((1, p[1].right))
        stack.append((0, p[1]))
        stack.append((1, p[1].left))
    return True

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isValidBST(root *TreeNode) bool {
	var prev *TreeNode

	var dfs func(root *TreeNode) bool
	dfs = func(root *TreeNode) bool {
		if root == nil {
			return true
		}
		if !dfs(root.Left) {
			return false
		}
		if prev != nil && prev.Val >= root.Val {
			return false
		}
		prev = root
		if !dfs(root.Right) {
			return false
		}
		return true
	}

	return dfs(root)
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isValidBST = function (root) {
    let prev = null;

    let dfs = function (root) {
        if (!root) {
            return true;
        }
        if (!dfs(root.left)) {
            return false;
        }
        if (prev && prev.val >= root.val) {
            return false;
        }
        prev = root;
        if (!dfs(root.right)) {
            return false;
        }
        return true;
    };

    return dfs(root);
};

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    private TreeNode prev;

    public bool IsValidBST(TreeNode root) {
        prev = null;
        return dfs(root);
    }

    private bool dfs(TreeNode root) {
        if (root == null)
        {
            return true;
        }
        if (!dfs(root.left))
        {
            return false;
        }
        if (prev != null && prev.val >= root.val)
        {
            return false;
        }
        prev = root;
        if (!dfs(root.right))
        {
            return false;
        }
        return true;
    }
}

98 - Validate Binary Search Tree

Question

Algorithm

Note: remember below in invalid, 7 should go right of 5

Code

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Question

Algorithm

Note: remember below in invalid, 7 should go right of 5

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