[LeetCode]109. Convert Sorted List to Binary Search Tree

[LeetCode]109. Convert Sorted List to Binary Search Tree

题目描述

思路

和 [LeetCode]108. Convert Sorted Array to Binary Search Tree 问题思路相同，只是数据结构从数组变成了链表，找到链表重点的方法，需要用快慢指针

代码

#include <iostream>
#include <vector>

using namespace std;

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(NULL) {}
};

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    ListNode* vectorToList(vector<int> nums) {
        if (nums.size() == 0)
            return NULL;
        ListNode *head = new ListNode(nums[0]),
            *iter = head;
        for (int i = 1; i < nums.size(); ++i) {
            ListNode *temp = new ListNode(nums[i]);
            iter->next = temp;
            iter = iter->next;
        }
        return head;
    }

    TreeNode* sortedListToBST(ListNode* head) {
        return insert(head, NULL);
    }

    TreeNode* insert(ListNode* head, ListNode* tail) {
        if (head == NULL || head == tail || (tail && tail->next == head))
            return NULL;
        ListNode *slow = head, *fast = head;
        while (fast != tail) {
            if (fast == tail || fast->next == tail)
                break;
            slow = slow->next;
            fast = fast->next->next;
        }
        TreeNode *root = new TreeNode(slow->val);

        root->left = insert(head, slow);
        root->right = insert(slow->next, tail);
        return root;
    }

    void visit(TreeNode* root) {
        if (root == NULL)
            return;
        if (root->left)
            visit(root->left);
        cout << root->val << " ";
        if (root->right)
            visit(root->right);
    }
};

int main() {
    vector<int> nums = { 1,2,3,4,5,6,7 };
    Solution s;
    ListNode *head = s.vectorToList(nums);
    TreeNode *root = s.sortedListToBST(head);
    s.visit(root);
    cout << endl;

    system("pause");
    return 0;
}