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:

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode *sortedArrayToBST(vector<int> &num) {
13         // Note: The Solution object is instantiated only once and is reused by each test case.
14         if(num.size() <= 0){
15             return nullptr;
16         }else{
17             return constructBSTFromArray(num, 0, (int)(num.size() - 1));
18         }
19     }
20 private:
21     TreeNode *constructBSTFromArray(vector<int> &num, int left, int right) {
22         int mid;
23         TreeNode* root;
24         
25         if(left > right){
26             return nullptr;
27         }
28         
29         mid = (right - left) / 2 + left;
30         root = new TreeNode(num[mid]);
31         root->left = constructBSTFromArray(num, left, mid - 1);
32         root->right = constructBSTFromArray(num, mid + 1, right);
33         
34         return root;
35     }
36 };

 

 posted on 2014-01-08 02:39  zhuli19901106  阅读(162)  评论(0编辑  收藏  举报