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leetcode No108. Convert Sorted Array to Binary Search Tree

时间:2016-08-24 09:56:49      阅读:190      评论:0      收藏:0      [点我收藏+]

Question:

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

把有序数组转化成平衡的BST

Algorithm:

找到数组中间的元素,作为根节点,则根节点左边是左子树,根节点右边是右子树,接着递归

Accepted Code:

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



leetcode No108. Convert Sorted Array to Binary Search Tree

原文:http://blog.csdn.net/u011391629/article/details/52296858

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