C++实现LeetCode(109.将有序链表转为二叉搜索树)

[leetcode] 109.convert sorted list to binary search tree 将有序链表转为二叉搜索树

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

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.

example:

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

这道题是要求把有序链表转为二叉搜索树，和之前那道 convert sorted array to binary search tree 思路完全一样，只不过是操作的数据类型有所差别，一个是数组，一个是链表。数组方便就方便在可以通过index直接访问任意一个元素，而链表不行。由于二分查找法每次需要找到中点，而链表的查找中间点可以通过快慢指针来操作，可参见之前的两篇博客 reorder list 和 linked list cycle ii 有关快慢指针的应用。找到中点后，要以中点的值建立一个数的根节点，然后需要把原链表断开，分为前后两个链表，都不能包含原中节点，然后再分别对这两个链表递归调用原函数，分别连上左右子节点即可。代码如下：

解法一：

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;
    }
};

我们也可以采用如下的递归方法，重写一个递归函数，有两个输入参数，子链表的起点和终点，因为知道了这两个点，链表的范围就可以确定了，而直接将中间部分转换为二叉搜索树即可，递归函数中的内容跟上面解法中的极其相似，参见代码如下：

解法二：

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;
    }
};

类似题目：

convert sorted array to binary search tree

参考资料：

https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/discuss/35476/share-my-java-solution-1ms-very-short-and-concise.

https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/discuss/35470/recursive-bst-construction-using-slow-fast-traversal-on-linked-list

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