Leetcode解题: Convert Sorted Array to Binary Search Tree (108)

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

解法1： Recursion

很多Tree的问题都可以用Divide & Conquer/递归的思想。这题也如此。要建立balanced tree，我们需要左边和右边的height尽可能相近。就考虑到选择中间作为root。之后就转化为把左array和右array转换的问题，这就是一个递归。

C++

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

Java

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