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