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108. Convert Sorted Array to Binary Search Tree

时间:2018-01-08 20:14:29      阅读:122      评论:0      收藏:0      [点我收藏+]

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108. Convert Sorted Array to Binary Search Tree

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

解析


- vector的初始化方法,递归创建左右子树的思想!

// Convert Sorted Array to Binary Search Tree
class Solution_108 {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        if (nums.size()==0)
        {
            return NULL;
        }
        if (nums.size()==1)
        {
            TreeNode* temp = new TreeNode(nums[0]);
            return temp;
        }

        int mid = nums.size() / 2;

        TreeNode* root = new TreeNode(nums[mid]);

        auto leftTree = vector<int>(nums.begin(), nums.begin() + mid);//最后一个迭代器指向最后一个元素的下一个位置
        auto rightTree = vector<int>(nums.begin() + mid + 1, nums.end());

        root->left = sortedArrayToBST(leftTree);
        root->right = sortedArrayToBST(rightTree);
        
        return root;
    }
};

108. Convert Sorted Array to Binary Search Tree

标签:source   oss   dep   递归   迭代器   present   位置   coder   post   

原文地址:https://www.cnblogs.com/ranjiewen/p/8244603.html

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