[LeetCode] Convert Sorted List to Binary Search Tree 將有序鏈表轉爲二叉搜索樹

時間 2019-11-14

標籤 leetcode convert sorted list binary search tree 有序鏈表轉爲搜索简体版

原文原文鏈接

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.node

Example:數組

Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

這道題是要求把有序鏈表轉爲二叉搜索樹，和以前那道 Convert Sorted Array to Binary Search Tree 思路徹底同樣，只不過是操做的數據類型有所差異，一個是數組，一個是鏈表。數組方便就方便在能夠經過index直接訪問任意一個元素，而鏈表不行。因爲二分查找法每次須要找到中點，而鏈表的查找中間點能夠經過快慢指針來操做，可參見以前的兩篇博客 Reorder List 和 Linked List Cycle II 有關快慢指針的應用。找到中點後，要以中點的值創建一個數的根節點，而後須要把原鏈表斷開，分爲先後兩個鏈表，都不能包含原中節點，而後再分別對這兩個鏈表遞歸調用原函數，分別連上左右子節點便可。代碼以下：

函數

解法一：post

class Solution {
public:
    TreeNode *sortedListToBST(ListNode* head) {
        if (!head) return NULL;
        if (!head->next) return new TreeNode(head->val);
        ListNode *slow = head, *fast = head, *last = slow;
        while (fast->next && fast->next->next) {
            last = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        fast = slow->next;
        last->next = NULL;
        TreeNode *cur = new TreeNode(slow->val);
        if (head != slow) cur->left = sortedListToBST(head);
        cur->right = sortedListToBST(fast);
        return cur;
    }
};

咱們也能夠採用以下的遞歸方法，重寫一個遞歸函數，有兩個輸入參數，子鏈表的起點和終點，由於知道了這兩個點，鏈表的範圍就能夠肯定了，而直接將中間部分轉換爲二叉搜索樹便可，遞歸函數中的內容跟上面解法中的極其類似，參見代碼以下：this

解法二：url

class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        if (!head) return NULL;
        return helper(head, NULL);
    }
    TreeNode* helper(ListNode* head, ListNode* tail) {
        if (head == tail) return NULL;
        ListNode *slow = head, *fast = head;
        while (fast != tail && fast->next != tail) {
            slow = slow->next;
            fast = fast->next->next;
        }
        TreeNode *cur = new TreeNode(slow->val);
        cur->left = helper(head, slow);
        cur->right = helper(slow->next, tail);
        return cur;
    }
};