LeetCode 108: Convert Sorted Array to Binary Search Tree DFS求解

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

 

解题思路：

1. 找到数组的中间节点将其置为根节点

2. 左边的即为左子树

3. 右边的即为右子树

4. 递归求解

 

/**
 * 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* sortBST(vector<int>& nums, int begin, int end){
        if(begin > end)
            return 0;
        int rootIndex = begin+(end-begin)/2;    //更简单的应该是    (begin+end)/2
        TreeNode* root = new TreeNode(nums[rootIndex]);
        root->left = sortBST(nums, begin, rootIndex-1);
        root->right = sortBST(nums, rootIndex+1, end);
        return root;
    }
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return sortBST(nums, 0, nums.size()-1);
    }
};