[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.

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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;
}