【32】98. Validate Binary Search Tree

98. Validate Binary Search Tree

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• Total Submissions: 620671
• Difficulty: Medium
• Contributors: Admin

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

Binary tree [2,1,3], return true.

Example 2:

    1
   / \
  2   3

Binary tree [1,2,3], return false.

Solution 1: Recursion without inorder

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
//Recursion without inorder
    bool isValidBST(TreeNode* root) {
        return helper(root, LONG_MAX, LONG_MIN);
    }
    
    bool helper(TreeNode* root, long left, long right){
        if(!root) return true;
        if(root -> val >= left || root -> val <= right) return false;//base case
        return helper(root -> right, left, root -> val) && helper(root -> left, root -> val, right);
    }
};

Solution 2: Inorder

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
//Recursion with inorder
    bool isValidBST(TreeNode* root) {
        if(!root) return true;
        vector<int> in;
        inorder(root, in);
        for(int i = 1; i < in.size(); i++){
            if(in[i] <= in[i - 1]) return false;
        }
        return true;
    }
    
    void inorder(TreeNode* root, vector<int> & in){
        if(root -> left) inorder(root -> left, in);
        in.push_back(root -> val);
        if(root -> right) inorder(root -> right, in);
    }
};