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     

"""
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：

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