Convert Sorted Array to Binary Search Tree

https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/description/

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

Thoughts

平衡二叉搜索树也就是把中点作为root，左右分别对应左右子树。 因此利用树的递归定义，把原问题分成左右两个相同的子问题, 再把它们用中点合并起来即可。

Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private TreeNode helper(int start, int end, int[] nums) {
        if (start > end) {
            return null;
        }
        int mid = start + (end - start) / 2;
        TreeNode node = new TreeNode(nums[mid]);
        node.left = helper(start, mid - 1, nums);
        node.right = helper(mid + 1, end, nums);

        return node;
    }


    public TreeNode sortedArrayToBST(int[] nums) {
        if (nums == null || nums.length == 0) {
            return null;
        }

        return helper(0, nums.length - 1, nums);
    }
}

Analysis

数组每个节点访问一遍，因此时间复杂度为O(n)

Ver.2

用另一个 stack记录每个节点对应的range, 并且让结点先创建出来给left和right再修改值. 这样就不用费心去找left和right到底在哪了。

/**
 * 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) {
        const int N = nums.size();
        if (N == 0) return NULL;
        stack<TreeNode*> s;
        stack<pair<int, int>> ranges;
        const auto root = new TreeNode(nums[(N - 1) / 2]);
        s.push(root);
        ranges.push({0, N - 1});
        while (!s.empty()) {
            auto cur = s.top();
            const auto range = ranges.top();
            s.pop();
            ranges.pop();
            const int c_mid = range.first + (range.second - range.first) / 2; 
            cur->val = nums[c_mid];
            if (range.first != c_mid) {
                ranges.push({range.first, c_mid - 1});
                const auto l = new TreeNode(-1);
                cur->left = l;
                s.push(l);
            } 
            if (range.second != c_mid) {
                ranges.push({c_mid + 1, range.second});
                const auto r = new TreeNode(-1);
                cur->right = r;
                s.push(r);
            }
        }
        return root;
    }
};