109.Convert Sorted List to Binary Search Tree Leetcode Python

本文详细介绍了如何将一个有序链表转换为一个高度平衡的二叉搜索树,通过快慢指针寻找中点实现链表到树的高效转换。

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Convert Sorted List to Binary Search Tree Total Accepted: 32343 Total Submissions: 117376 My Submissions Question Solution 

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


这题和convert array to binary search tree不同点在与LinkedList 没有random access所以要想找到中点只能利用fast 和slow pointer来查找。

这种方法比直接把linkedlist转成array要慢。


代码如下:

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param head, a list node
    # @return a tree node
    def convert(self,head,tail):
        if head==tail:
            return None
        if head.next==tail:
            return TreeNode(head.val)
        mid=head
        fast=head
        while fast!=tail and fast.next!=tail:
            fast=fast.next.next
            mid=mid.next
        node=TreeNode(mid.val)
        node.left=self.convert(head,mid)
        node.right=self.convert(mid.next,tail)
        return node
    def sortedListToBST(self, head):
        return self.convert(head,None)
        


【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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