LeetCodeConvert Sorted List to Binary Search Tree

本文详细介绍了如何将链表转化为二叉搜索树的方法,包括链表的复制、中间节点的查找以及递归实现的过程。
 /*这道题与Convert Sorted Array to Binary Search Tree很类似,可以采用
 相同的方法解题,但是链表不支持随机访问,所以每次都需要查找链表的中间节点。
 同时,还有一个需要注意的地方:需要将输入的链表复制出来处理,否则会报错。*/
class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        if(head == nullptr) return nullptr;
        //复制原来的链表到新的链表中,因为不能改变原来的链表。
        //(试过不把原来的链表复制出来,但提交时失败的。)
        ListNode node(-1), *tail(&node);
        while(head != nullptr){
            tail->next = new ListNode(head->val);
            head = head->next;
            tail = tail->next;
        } 
        head = node.next;
        
        ListNode *fast(head), *slow(head), *pre_slow(nullptr);
        while(fast->next != nullptr && fast->next->next != nullptr){
            fast = fast->next->next;
            pre_slow = slow;
            slow = slow->next;
        }
        TreeNode *root = new TreeNode(slow->val);
        ListNode *p(slow->next);
        delete slow;
        if(pre_slow != nullptr){
            pre_slow->next = nullptr;
            root->left = sortedListToBST(head);   
        }
        root->right = sortedListToBST(p);
        return root;
    }
};

【Solution】 To convert a binary search tree into a sorted circular doubly linked list, we can use the following steps: 1. Inorder traversal of the binary search tree to get the elements in sorted order. 2. Create a doubly linked list and add the elements from the inorder traversal to it. 3. Make the list circular by connecting the head and tail nodes. 4. Return the head node of the circular doubly linked list. Here's the Python code for the solution: ``` class Node: def __init__(self, val): self.val = val self.prev = None self.next = None def tree_to_doubly_list(root): if not root: return None stack = [] cur = root head = None prev = None while cur or stack: while cur: stack.append(cur) cur = cur.left cur = stack.pop() if not head: head = cur if prev: prev.right = cur cur.left = prev prev = cur cur = cur.right head.left = prev prev.right = head return head ``` To verify the accuracy of the code, we can use the following test cases: ``` # Test case 1 # Input: [4,2,5,1,3] # Output: # Binary search tree: # 4 # / \ # 2 5 # / \ # 1 3 # Doubly linked list: 1 <-> 2 <-> 3 <-> 4 <-> 5 # Doubly linked list in reverse order: 5 <-> 4 <-> 3 <-> 2 <-> 1 root = Node(4) root.left = Node(2) root.right = Node(5) root.left.left = Node(1) root.left.right = Node(3) head = tree_to_doubly_list(root) print("Binary search tree:") print_tree(root) print("Doubly linked list:") print_list(head) print("Doubly linked list in reverse order:") print_list_reverse(head) # Test case 2 # Input: [2,1,3] # Output: # Binary search tree: # 2 # / \ # 1 3 # Doubly linked list: 1 <-> 2 <-> 3 # Doubly linked list in reverse order: 3 <-> 2 <-> 1 root = Node(2) root.left = Node(1) root.right = Node(3) head = tree_to_doubly_list(root) print("Binary search tree:") print_tree(root) print("Doubly linked list:") print_list(head) print("Doubly linked list in reverse order:") print_list_reverse(head) ``` The output of the test cases should match the expected output as commented in the code.
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