LeetCode #109 - Convert Sorted List to Binary Search Tree

最新推荐文章于 2021-11-02 12:49:42 发布

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本文介绍了一种将有序链表转换为高度平衡二叉搜索树的方法。通过使用快慢指针找到链表中点作为根节点，并递归地构造左右子树，最终形成一棵平衡的二叉搜索树。

题目描述：

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

与运用有序数组构造BST类似，但是确定中点需要利用快慢指针遍历才可以。

class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        if(head==NULL) return NULL;
        if(head->next==NULL)
        {
            TreeNode* root=new TreeNode(head->val);
            return root;
        }
        ListNode* p=head;
        ListNode* q=head;
        ListNode* pre_p=new ListNode(0);
        pre_p->next=p;
        while(q!=NULL&&q->next!=NULL)
        {
            pre_p=p;
            q=q->next->next;
            p=p->next;
        }
        pre_p->next=NULL;
        TreeNode* left=sortedListToBST(head);
        TreeNode* right=sortedListToBST(p->next);
        TreeNode* root=new TreeNode(p->val);
        root->left=left;
        root->right=right;
        return root;
    }
};