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Topic: 109. Binary search tree conversion from ordered linked list
Given a singly linked list, the elements in it are sorted in ascending order and converted to a highly balanced binary search tree.
In this question, a highly balanced binary tree means that the absolute value of the height difference between the left and right subtrees of each node of a binary tree does not exceed 1.
Example:
给定的有序链表： [-10, -3, 0, 5, 9],
一个可能的答案是：[0, -3, 9, -10, null, 5], 它可以表示下面这个高度平衡二叉搜索树：
0
/ \
-3 9
/ /
-10 5
Source: LeetCode (LeetCode)
Link: https://leetcode-cn.com/problems/convert-sorted-list-to-binary-search-tree The
copyright is owned by LeetCode . For commercial reprints, please contact the official authorization. For non-commercial reprints, please indicate the source.
Basic idea: first order traversal recursive construction
- Take the intermediate node of the linked list as the root, recursively construct the left and right subtrees
- Note: When looking for an intermediate node, you need to disconnect the intermediate node from the node in front of the intermediate node; and you need to pay attention to determine whether there is a left subtree
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* sortedListToBST(ListNode* head) {
if(head == NULL)
return NULL;
// if(head->next == NULL)
// return new TreeNode(head->val);
ListNode * mid;
mid = midNode(head);
ListNode * lhead = head, * rhead = mid->next;
TreeNode * root = new TreeNode(mid->val);
if(mid != head)//判断是否有左子树
root->left = sortedListToBST(lhead);
root->right = sortedListToBST(rhead);
return root;
}
ListNode* midNode(ListNode* head){
ListNode * fast = head, * slow = head, *pre = NULL;
while(fast->next != NULL && fast->next->next != NULL){
pre = slow;
slow = slow->next;
fast = fast->next->next;
}
if(pre)//将中间节点和前面链表断开
pre->next = NULL;
return slow;
}
};