[Leetcode 32] 108 Convert Sorted Array to Binary Search Tree

Problem:

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

 

Analysis:

Remember a binary search tree's inorder treversal is a sorted array. 

To convert in the contrary direction, we need to create a node with the middle value of the given array and then create the left treenode with the left part of the array and the right treenode with the right part of the array. Do it recursively!

 

Code:

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *sortedArrayToBST(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        return sortedArrayToBSTHelper(num, 0, num.size()-1);
    }
    
    TreeNode *sortedArrayToBSTHelper(vector<int> &num, int lo, int hi) {
        if (lo > hi)
            return NULL;
        
        int mid = (lo + hi) / 2;
        TreeNode *node = new TreeNode(num[mid]);
        node->left = sortedArrayToBSTHelper(num, lo, mid-1);
        node->right = sortedArrayToBSTHelper(num, mid+1, hi);
        
        return node;
    }
};

 

Attention:

To create an object in C++, either the following two ways:

1. SomeClass obj(para);

2. SomeClass *obj = new SomeClass(para);

The 1st will return an object, while the 2nd will return a pointer.