[leetcode] BST與鏈表的相互轉換

1. 鏈表-->BST (借鑑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;
    }
};

這樣做的好處是：不需要不斷地traverse，以尋找中間點！！！！！

2. BST-->鏈表

二叉樹轉換爲雙向鏈表的思想： 保存兩個關鍵的指針：pre指針(上一次訪問的指針)和T指針；對於每一個節點，執行T->lchild = prepre;pre->rchild = T; 然後執行pre = T;當中序遍歷結束時，二叉搜索樹也被轉化爲雙鏈表了。

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