lc257. Binary Tree Paths

weixin_34056162

于 2019-06-20 06:35:06 发布

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文章标签： python

原文链接：https://juejin.im/post/5d0b28d9f265da1ba431f363

博客围绕二叉树展开，给定一个二叉树，要求返回所有根到叶的路径。采用深度优先搜索（DFS）记录每条路径的数组来解决问题，并给出了Python3代码实现。

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Binary Tree Paths Easy

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

Note: A leaf is a node with no children.

Example:

Input:

1 /
2 3
5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

思路：dfs 记录每条线的数组代码：python3

class Solution:
    def binaryTreePaths(self, root: TreeNode) -> List[str]:
        if not root: return []
        arr=[]
        def dfs(root,ar):
            subAr=ar[:]
            if not root:return
            if not root.left and not root.right:
                subAr.append(str(root.val))
                arr.append(subAr)
            else:
                subAr.append(str(root.val))            
                dfs(root.left,subAr)
                dfs(root.right,subAr)                
        dfs(root,[])
        result=[]
        for a in arr:
            result.append('->'.join(a))
        return result
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