目录

一、题目内容

二、解题思路

三、代码

一、题目内容

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。

示例:

输入:

   1
 /   \
2     3
 \
  5

输出: ["1->2->5", "1->3"]

解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3

二、解题思路

回溯+DFS，简单易懂

三、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:
     
    
   

    

     

    
    

     
 def binaryTreePaths(self, root: TreeNode):
     
    
   

    

     

    
    

     

       paths = []
     
    
   

    

     

    
    

     
 if root is None:
     
    
   

    

     

    
    

     
 return []
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     
 def dfs(root, path):
     
    
   

    

     

    
    

     
 if root.left is None and root.right is None:
     
    
   

    

     

    
    

     

       paths.append(path)
     
    
   

    

     

    
    

     
 return
     
    
   

    

     

    
    

     
 if root.left is not None:
     
    
   

    

     

    
    

     

       org_path = path
     
    
   

    

     

    
    

     

       path += "->" + str(root.left.val)
     
    
   

    

     

    
    

     

       dfs(root.left, path)
     
    
   

    

     

    
    

     

       path = org_path
     
    
   

    

     

    
    

     
 if root.right is not None:
     
    
   

    

     

    
    

     

       org_path = path
     
    
   

    

     

    
    

     

       path += "->" + str(root.right.val)
     
    
   

    

     

    
    

     

       dfs(root.right, path)
     
    
   

    

     

    
    

     

       path = org_path
     
    
   

    

     

    
    

     

       dfs(root, str(root.val))
     
    
   

    

     

    
    

     
 return paths
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     

      if __name__ == '__main__':
     
    
   

    

     

    
    

     

       root = TreeNode(1)
     
    
   

    

     

    
    

     

       root.left = TreeNode(2)
     
    
   

    

     

    
    

     

       root.right = TreeNode(3)
     
    
   

    

     

    
    

     

       root.left.left = TreeNode(5)
     
    
   

    

     

    
    

     

       s = Solution()
     
    
   

    

     

    
    

     

       ans = s.binaryTreePaths(root=root)
     
    
   

    

     

    
    

     

       print(ans)
     
    
  
 

文章来源: nickhuang1996.blog.csdn.net，作者：悲恋花丶无心之人，版权归原作者所有，如需转载，请联系作者。

原文链接：nickhuang1996.blog.csdn.net/article/details/108400545