[leetcode] BST与链表的相互转换

1. 链表-->BST (借鉴smile的)

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; *//** * Definition 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int cnt = 0;        ListNode* cur = head;        while (cur)        {            cur = cur->next;            cnt++;        }                return build(head, 0, cnt - 1);    }        TreeNode* build(ListNode* &head, int left, int right)    {        if (left > right)            return NULL;                            int mid = left + (right - left) / 2;        TreeNode* leftChild = build(head, left, mid - 1);                            TreeNode* parent = new TreeNode(head->val);        parent->left = leftChild;        head = head->next;        parent->right = build(head, mid + 1, right);        return parent;    }};

这样做的好处是：不需要不断地traverse，以寻找中间点！！！！！

2. BST-->链表

二叉树转换为双向链表的思想： 保存两个关键的指针：pre指针(上一次访问的指针)和T指针；对于每一个节点，执行T->lchild = prepre;pre->rchild = T; 然后执行pre = T;当中序遍历结束时，二叉搜索树也被转化为双链表了。

void BiTreeToLinklist(BiTree &T)  // change tree to linklist {     if(T != NULL) // 双向链表化     {         BiTreeToLinklist(T->lchild);         T->lchild = pre;         pre->rchild = T;         pre = T;         BiTreeToLinklist(T->rchild);     } }