[LeetCode] 114. Flatten Binary Tree to Linked List

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本文介绍了一种将二叉树结构通过递归方式转换为链表的方法。具体思路是将二叉树的左子树放置到右子树的位置，并将原本的右子树连接到左子树的最右端节点上。文章提供了详细的实现代码。

题：https://leetcode.com/problems/flatten-binary-tree-to-linked-list/description/

题目

Given a binary tree, flatten it to a linked list in-place.

For example, given the following tree:

/ \
2 5
/ \ \
3 4 6
The flattened tree should look like:

1
\
2
\
3
\
4
\
5
\
6

思路

递归，把左树放在右树的位置，将右树放到，左树的末端。

code

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def flatten(self, root):
        """
        :type root: TreeNode
        :rtype: void Do not return anything, modify root in-place instead.
        """

        if root == None:
            return None
        leftnode = root.left
        rightnode = root.right
        root.left = None

        self.flatten(leftnode)
        self.flatten(rightnode)

        root.right = leftnode;

        cur = root
        while cur.right != None:
            cur = cur.right
        cur.right = rightnode