108 Convert Sorted Array to Binary Search Tree

1 题目

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

2 尝试解

2.1 分析

采用递归思路，如[1,2,3,4,5]，以3为根节点，左子树为[1,2]，右子树[4,5]，实现递归。递归需要四个参数，原始数组，待建立子树对应数组中的起点和终点，子树的根节点。

2.2 代码

class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        TreeNode* root = NULL; 
        if(nums.size() == 0) return root;
        root = new TreeNode(0);
        BuildTree(nums,0,nums.size()-1,root);
        return root;
    }
    void BuildTree(vector<int>& nums, int start,int end,TreeNode* root){
        if(start == end) {
            root->val = nums[start];
        }
        else{
            int mid = (start + end)/2;
            root->val = nums[mid];
            if(mid < end){
                root->right = new TreeNode(0);
                BuildTree(nums,mid+1,end,root->right);
            }
            if(start < mid){
                root->left = new TreeNode(0);
                BuildTree(nums,start,mid-1,root->left);
            }
        }   
    }
};

3 标准解

class Solution {
public:
    TreeNode *sortedArrayToBST(vector<int> &num) {
        if(num.size() == 0) return NULL;
        if(num.size() == 1)
        {
            return new TreeNode(num[0]);
        }
        
        int middle = num.size()/2;
        TreeNode* root = new TreeNode(num[middle]);
        
        vector<int> leftInts(num.begin(), num.begin()+middle);
        vector<int> rightInts(num.begin()+middle+1, num.end());
        
        root->left = sortedArrayToBST(leftInts);
        root->right = sortedArrayToBST(rightInts);
        
        return root;
    }
};