257. Binary Tree Paths [easy] (Python)

题目链接

https://leetcode.com/problems/binary-tree-paths/

题目原文

Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

   1
 /   \
2     3
 \
  5

All root-to-leaf paths are:
[“1->2->5”, “1->3”]

题目翻译

给定一个二叉树，返回所有从根到叶子的路径。

思路方法

思路一

递归实现深度优先遍历。注意要记录途中访问过的节点，遇到叶子节点时可以生成一条路径字符串。关于记录路径，一般的方法（后面几种思路的方法）是直接将目前为止的路径字符串记录下来，而下面的代码采用先保存在数组里需要生成路径时再生成的方法，大家可以对照比较一下。

代码

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

class Solution:
    # @param {TreeNode} root
    # @return {string[]}
    def binaryTreePaths(self, root):
        res, path_list = [], []
        self.dfs(root, path_list, res)
        return res

    def dfs(self, root, path_list, res):
        if not root:
            return
        path_list.append(str(root.val))
        if not root.left and not root.right:
            res.append('->'.join(path_list))
        if root.left:
            self.dfs(root.left, path_list, res)
        if root.right:
            self.dfs(root.right, path_list, res)
        path_list.pop()

思路二

上面的递归定义了一个额外的helper function——dfs()，用于传递和保存递归的信息。其实可以直接用原函数做递归，代码如下。

代码

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

class Solution:
    # @param {TreeNode} root
    # @return {string[]}
    def binaryTreePaths(self, root):
        res = []
        if not root:
            return res
        if not root.left and not root.right:
            res.append(str(root.val))
            return res
        for path in self.binaryTreePaths(root.left):
            res.append(str(root.val) + '->' + path)
        for path in self.binaryTreePaths(root.right):
            res.append(str(root.val) + '->' + path)
        return res

思路三

有递归DFS代码，就可以用栈实现DFS。

代码

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

class Solution:
    # @param {TreeNode} root
    # @return {string[]}
    def binaryTreePaths(self, root):
        res, stack = [], [(root, '')]
        while stack:
            node, curs = stack.pop()
            if node:
                if not node.left and not node.right:
                    res.append(curs + str(node.val))
                stack.append((node.left, curs + str(node.val) + '->'))
                stack.append((node.right, curs + str(node.val) + '->'))
        return res

思路四

同样的，用队列实现BFS也可以完成要求。

代码

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

class Solution:
    # @param {TreeNode} root
    # @return {string[]}
    def binaryTreePaths(self, root):
        res, queue = [], [(root, '')]
        while queue:
            node, curs = queue.pop()
            if node:
                if not node.left and not node.right:
                    res.append(curs + str(node.val))
                queue.insert(0, (node.left, curs + str(node.val) + '->'))
                queue.insert(0, (node.right, curs + str(node.val) + '->'))
        return res

PS: 新手刷LeetCode，新手写博客，写错了或者写的不清楚还请帮忙指出，谢谢！
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