98. Validate Binary Search Tree

98. Validate Binary Search Tree (Medium)

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:

    2
   / \
  1   3

Input: [2,1,3]
Output: true

Example 2:

    5
   / \
  1   4
     / \
    3   6

Input: [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.

Related Topics:
Tree, Depth-first Search

Similar Questions:

• Binary Tree Inorder Traversal (Medium)
• Find Mode in Binary Search Tree (Easy)

Solution 1. Pre-order Traversal

// OJ: https://leetcode.com/problems/validate-binary-search-tree
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
private:
  bool isValidBST(TreeNode *root, TreeNode *lb, TreeNode *rb) {
    if (!root) return true;
    if ((lb && root->val <= lb->val) || (rb && root->val >= rb->val)) return false;
    return isValidBST(root->left, lb, root) && isValidBST(root->right, root, rb);
  }
public:
  bool isValidBST(TreeNode* root) {
    return isValidBST(root, NULL, NULL);
  }
};

Or

// OJ: https://leetcode.com/problems/validate-binary-search-tree
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
public:
    bool isValidBST(TreeNode* root, TreeNode *left = NULL, TreeNode *right = NULL) {
        if (!root) return true;
        return (!left || root->val > left->val) && (!right || root->val < right->val) && isValidBST(root->left, left, root) && isValidBST(root->right, root, right);
    }
};

Solution 2. In-order traversal

// OJ: https://leetcode.com/problems/validate-binary-search-tree
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(logN)
class Solution {
private:
  TreeNode *last = NULL;
public:
  bool isValidBST(TreeNode* root) {
    if (!root) return true;
    if (!isValidBST(root->left) || (last && last->val >= root->val)) return false;
    last = root;
    return isValidBST(root->right);
  }
};