LeetCode

Convert Sorted Array to Binary Search Tree

2014.1.8 02:11

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

Solution:

　　A height balanced binary tree is a binary tree, whose heights of left and right subtrees varie by at most 1. Thus constructing such a tree from a sorted array can be done by cutting the array by half and picking the middle element as the root.

　　Note that the inorder traversal of a BST is a sorted array, that should help you figure out how the method works.

　　Do this procedure recursively and the job is done.

　　Time and space complexities are both O(n), where n is the number of elements in the array. Space complexity comes from local parameters in function calls.

Accepted 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(num.size() <= 0){
            return nullptr;
        }else{
            return constructBSTFromArray(num, 0, (int)(num.size() - 1));
        }
    }
private:
    TreeNode *constructBSTFromArray(vector<int> &num, int left, int right) {
        int mid;
        TreeNode* root;
        
        if(left > right){
            return nullptr;
        }
        
        mid = (right - left) / 2 + left;
        root = new TreeNode(num[mid]);
        root->left = constructBSTFromArray(num, left, mid - 1);
        root->right = constructBSTFromArray(num, mid + 1, right);
        
        return root;
    }
};