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108. Convert Sorted Array to Bin

108. Convert Sorted Array to Bin

作者: SilentDawn | 来源:发表于2018-06-18 10:39 被阅读0次

    Problem

    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
    

    Code

    /**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    static int var = [](){
        std::ios::sync_with_stdio(false);
        cin.tie(NULL);
        return 0;
    }();
    class Solution {
    public:
        TreeNode* sortedArrayToBST(vector<int>& nums) {
            return func(nums);
        }
        TreeNode* func(vector<int> nums){
            if(nums.empty())
                return NULL;
            int mid = nums.size()/2;
            TreeNode* root = new TreeNode(nums[mid]);
            vector<int> left_temp(nums.begin(),nums.begin()+mid);
            root->left = func(left_temp);
            vector<int> right_temp(nums.begin()+mid+1,nums.end());
            root->right = func(right_temp);
            return root;
        }
    };
    

    Result

    108. Convert Sorted Array to Binary Search Tree.png

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