【Leetcode】257—Binary Tree Paths

一、题目描述

给定一个二叉树，返回所有从根节点到叶子节点的路径。
说明: 叶子节点是指没有子节点的节点。

二、代码实现

实现一：基于栈

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

class Solution(object):
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        if not root: return []
        res = []
        stack = [(root, "")]
        while stack:
            node, ls = stack.pop()
            if not node.left and not node.right:
                res.append(ls + str(node.val))
            
            if node.left:
                stack.append((node.left, ls + str(node.val) + '->'))
            
            if node.right:
                stack.append((node.right, ls + str(node.val) + '->'))
        
        return res

实现二：DFS

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

class Solution(object):
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        def dfs(root, path, res):
            if not root.left and not root.right:
                res.append(path + str(root.val))
                return res
            if root.left:
                dfs(root.left, path + str(root.val) + "->", res)
            if root.right:
                dfs(root.right, path + str(root.val) + "->", res)
            
        if not root: return []
        
        res = [] 
        dfs(root, "", res)
        
        return res