【Leetcode每日笔记】114. 二叉树展开为链表(Python)

题目

给定一个二叉树，原地将它展开为一个单链表。

解题思路

前序遍历

递归实现前序遍历

def dfs(root):
	if not root:
		return 
	res.append(root.val)
	dfs(root.left)
	dfs(root.right)

迭代实现前序遍历

cur,stack,res = root,[],[]
while cur or stack:
	while cur:
		stack.append(cur)
		res.append(cur.val)
		cur = cur.left
	tmp = stack.pop()
	cur = tmp.right

寻找前驱节点

代码

class Solution:
    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        stack = []
        p = root
        res = []
        while p or stack:
            while p:
                stack.append(p)
                res.append(p.val)
                p = p.left
            tmp = stack.pop()
            p = tmp.right
        res = res[1:]
        while res:
            root.left = None
            root.right = TreeNode(res.pop(0))
            root = root.right

class Solution:
    def flatten(self, root: TreeNode) -> None:
        curr = root
        while curr:
            if curr.left:
                predecessor = nxt = curr.left
                while predecessor.right:
                    predecessor = predecessor.right
                predecessor.right = curr.right
                curr.left = None
                curr.right = nxt
            curr = curr.right