Leetcode-108. Convert Sorted Array to Binary Search Tree

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5
           

思路：此题仅仅是平衡二叉树，因此不需要考虑数值大小，只需要调整根节点的两个分支数量即可。每一次都选取数组最中间的数值作为之前选取节点的根节点，递归操作，就可以保证二叉树的平衡。

/**
 * 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:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        if(nums.size()==0)
            return NULL;
        if(nums.size()==1)
            return new TreeNode(nums[0]);
        
        int mid = nums.size()/2;
        TreeNode* root = new TreeNode(nums[mid]);
        
        vector<int> leftlist(nums.begin(),nums.begin()+mid);
        vector<int> rightlist(nums.begin()+mid+1,nums.end());
        
        root->left = sortedArrayToBST(leftlist);
        root->right = sortedArrayToBST(rightlist);
        
        return root;
    }
};