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微软面试模拟题 Leetcode 108. 将有序数组转换为二叉搜索树
阅读量:229 次
发布时间:2019-03-01

本文共 463 字,大约阅读时间需要 1 分钟。

每次取中点作为根节点,递归的处理,就能保住构建出来的二叉树一定是平衡二叉树。

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

 

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