算法 dfs 二叉树的所有路径

480. 二叉树的所有路径

给一棵二叉树，找出从根节点到叶子节点的所有路径。

Example

样例 1:

输入：{1,2,3,#,5}
输出：["1->2->5","1->3"]
解释：
   1
 /   \
2     3
 \
  5

样例 2:

输入：{1,2}
输出：["1->2"]
解释：
   1
 /   
2     

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""" Definition of TreeNode: class TreeNode:     def __init__(self, val):         self.val = val         self.left, self.right = None, None """   class Solution:     """     @param root: the root of the binary tree     @return: all root-to-leaf paths     """     def binaryTreePaths(self, root):         # write your code here         path = []         result = []         self.dfs(root, path, result)         return result           def dfs(self, root, path, result):         if not root:             return                   path.append(str(root.val))         if root.left is None and root.right is None:             result.append("->".join(path))         else:                        self.dfs(root.left, path, result)             self.dfs(root.right, path, result)         path.pop()

　　

还有使用分治实现的，jiuzhang给的：

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class Solution:     """     @param root: the root of the binary tree     @return: all root-to-leaf paths     """     def binaryTreePaths(self, root):         if root is None:             return []                       if root.left is None and root.right is None:             return [str(root.val)]           leftPaths = self.binaryTreePaths(root.left)         rightPaths = self.binaryTreePaths(root.right)                   paths = []         for path in leftPaths + rightPaths:             paths.append(str(root.val) + '->' + path)                       return paths