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98. 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

Solutions

Solution 1: Recursion

In-order traversal. If it is a valid binary search tree, then the sequence traversed should be monotonically increasing. So, we only need to compare and judge whether the current number traversed is greater than the previous number.

Alternatively, consider the subtree with root as the root, whether all node values are within the valid range, and judge recursively.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the tree.

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

   5
  / \
 1   8
  \   \
   7   10

  • /**
 * 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;
    }
}

  • /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* prev;

    bool isValidBST(TreeNode* root) {
        prev = nullptr;
        return dfs(root);
    }

    bool dfs(TreeNode* 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;
    }
};

  • # 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: # recursive
    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:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isValidBST(self, root: TreeNode) -> bool:
        def dfs(root):
            nonlocal prev
            if root is None:
                return True
            if not dfs(root.left):
                return False
            if prev >= root.val:
                return False
            prev = root.val
            if not dfs(root.right):
                return False
            return True

        prev = -inf
        return dfs(root)

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

# 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;
    }
}

  • /**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function isValidBST(root: TreeNode | null): boolean {
    let prev: TreeNode | null = null;
    const dfs = (root: TreeNode | null): boolean => {
        if (!root) {
            return true;
        }
        if (!dfs(root.left)) {
            return false;
        }
        if (prev && prev.val >= root.val) {
            return false;
        }
        prev = root;
        return dfs(root.right);
    };
    return dfs(root);
}


  • // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, prev: &mut Option<i32>) -> bool {
        if root.is_none() {
            return true;
        }
        let root = root.as_ref().unwrap().borrow();
        if !Self::dfs(&root.left, prev) {
            return false;
        }
        if prev.is_some() && prev.unwrap() >= root.val {
            return false;
        }
        *prev = Some(root.val);
        Self::dfs(&root.right, prev)
    }

    pub fn is_valid_bst(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
        Self::dfs(&root, &mut None)
    }
}

All Problems

All Solutions