LeetCode 109. Convert Sorted List to Binary Search Tree

LeetCode 109. Convert Sorted List to Binary Search Tree

Description:

Given a singly linked list 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.

Example:

Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0     / \   -3   9   /   / -10  5

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分析：

这道题可以参考108题将数组转换成二叉搜索树的代码，我也是利用它来完成这道题目。

首先遍历链表，将其链表值存入vector数组中，然后和108题一样将数组转换成二叉搜索树。
具体过程为：找出中间数，作为根节点，然后将数组分成两半，递归求解即可。

LeetCode上的Discuss有一个想法是将找出链表的中间节点：先初始化temp为head，然后用了一个循环，在循环里不断更新temp为temp = temp->next->next;，循环退出条件为temp != NULL && temp->next != NULL，得到链表的中间节点，然后递归左右链表。具体代码在文章底部。

代码如下：

// Definition for singly-linked list.struct ListNode {    int val;    ListNode *next;    ListNode(int x) : val(x), next(NULL) {}};/** * 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* sortedListToBST(ListNode* head) {        if (head == NULL) return NULL;        if (head->next == NULL) return new TreeNode(head->val);        int cnt = 0;        vector<int> v;        ListNode* temp = head;        while (temp != NULL) {            v.push_back(temp->val);            temp = temp->next;            cnt++;        }        return sortedArrayToBST(v);    }    TreeNode* sortedArrayToBST(vector<int>& nums) {        if (nums.size() == 0) return NULL;        if (nums.size() == 1) return new TreeNode(nums[0]);        // 根节点        int middle = nums.size() / 2;        TreeNode* root = new TreeNode(nums[middle]);        // 左子树/右子树        vector<int> leftVectors(nums.begin(), nums.begin() + middle);        vector<int> rightVectors(nums.begin() + middle + 1, nums.end());        // 递归        root->left = sortedArrayToBST(leftVectors);        root->right = sortedArrayToBST(rightVectors);        return root;    }};

Discuss:

#include <iostream>#include <vector>#include <malloc.h>using namespace std;// Definition for singly-linked list.struct ListNode {    int val;    ListNode *next;    ListNode(int x) : val(x), next(NULL) {}};/** * 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* sortedListToBST(ListNode* head) {        return sortedListToBST(head, NULL);    }private:    TreeNode* sortedListToBST(ListNode* head, ListNode* tail) {        if (head == tail) return NULL;        if (head->next == tail) {            TreeNode *root = new TreeNode(head->val);            return root;        }        ListNode *mid = head, *temp = head;        // 寻找中间节点        while (temp != tail && temp->next != tail) {            mid = mid->next;            temp = temp->next->next;        }        TreeNode *root = new TreeNode(mid->val);        root->left = sortedListToBST(head, mid);        root->right = sortedListToBST(mid->next, tail);        return root;    }};