LeetCode

Validate Binary Search Tree

2013.12.31 18:48

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / 
 2   3
    /
   4
    
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

Solution:

　　The inorder traversal of a BST is a sorted array, that's how my code verified the tree.

　　Perform an inorder traversal and see if the result is sorted. If so, valid; otherwise, no. An extra array is needed to hold the inorder traversal result.

　　Time complexity and space complexity are both O(n), where n is the number of nodes in a tree.

Accepted code:

// 1WA, 1CE, 1AC
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(root == nullptr){
            return true;
        }
        arr.clear();
        // 1WA here, the in-order traversal of BST is sorted.
        inorderTraversal(root);
        
        int i, n;
        
        n = arr.size();
        for(i = 0; i < n - 1; ++i){
            if(arr[i] >= arr[i + 1]){
                break;
            }
        }
        arr.clear();
        
        return (i == n - 1);
    }
private:
    vector<int> arr;
    
    void inorderTraversal(TreeNode *root) {
        // 1CE, null or nullptr...
        if(root == nullptr){
            return;
        }
        
        if(root->left != nullptr){
            inorderTraversal(root->left);
        }
        arr.push_back(root->val);
        if(root->right != nullptr){
            inorderTraversal(root->right);
        }
    }
};