[LeetCode] Convert Sorted List to Binary Search Tree DFS,深度搜索

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

 

Hide Tags

  Depth-first Search Linked List

 

---

    这题是将链表变成二叉树，比较麻烦的遍历过程，因为链表的限制，所以深度搜索的顺序恰巧是链表的顺序，通过设置好递归函数的参数，可以在深度搜索时候便可以遍历了。

 

TreeNode * help_f(ListNode *&curList,int lft,int rgt)

 

全部代码：

#include <iostream>

using namespace std;



/**

 * Definition for singly-linked list.

 */

struct ListNode {

    int val;

    ListNode *next;

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

};



/**

 * Definitiosn for binary tree

 */

struct TreeNode {

    int val;

    TreeNode *left;

    TreeNode *right;

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

};



class Solution {

public:

    TreeNode *sortedListToBST(ListNode *head) {

        int len=0;

        ListNode * p = head;

        while(p!=NULL){

            len++;

            p=p->next;

        }

//        cout<<len<<endl;

        return help_f(head,0,len-1);

    }



    TreeNode * help_f(ListNode *&curList,int lft,int rgt)

    {

        if(lft>rgt) return  NULL;

        int mid=(lft+rgt)/2;

        TreeNode *lftCld = help_f(curList,lft,mid-1);

        TreeNode *parent =new TreeNode(curList->val);

        parent->left=lftCld;

        curList=curList->next;

        parent->right=help_f(curList,mid+1,rgt);

        return parent;

    }

};



int main()

{

    ListNode n1(0);

    Solution sol;

    sol.sortedListToBST(&n1);

    return 0;

}