Binary Tree Paths（leetcode 257）

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

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

# 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):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        def construct_paths(root, path):
            if root:
                path += str(root.val)
                if not root.left and not root.right:  # 当前节点是叶子节点
                    paths.append(path)  # 把路径加入到答案中
                else:
                    path += '->'  # 当前节点不是叶子节点，继续递归遍历
                    construct_paths(root.left, path)
                    construct_paths(root.right, path)

        paths = []
        construct_paths(root, '')
        return paths