【easy】108. Convert Sorted Array to Binary Search Tree

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.

//思路还是很清晰的：就是要取中间节点作为根节点，其他的分别是左子树和右子树，递归
/**
 * 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) {
        TreeNode *node = createBST(nums,0,nums.size()-1);
        return node;
    }
public:
    TreeNode* createBST(vector<int>nums, int left, int right){
        if (left>right)
            return NULL;//??
        int mid = (left+right)/2;
        TreeNode *node = new TreeNode(nums[mid]);
        node->left = createBST(nums, left, mid - 1);
        node->right = createBST(nums, mid+1, right);
        return node;
    }
};

递归出口NULL注意，最终只是要返回根节点。

原文地址：https://www.cnblogs.com/sherry-yang/p/8407397.html