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。
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: TreeNode) -> List[str]:
if not root:
return []
if not root.left and not root.right:
return [str(root.val)]
paths = self.binaryTreePaths(root.left)
paths.extend(self.binaryTreePaths(root.right))
return ["%d->" % root.val + path for path in paths]
# 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) -> List[str]:
if not root:
return []
paths = []
nodes = [(root, str(root.val))]
while nodes:
curr, path = nodes.pop()
if not curr.left and not curr.right:
paths.append(path)
if curr.left:
nodes.append((curr.left, path + "->" + str(curr.left.val)))
if curr.right:
nodes.append((curr.right, path + "->" + str(curr.right.val)))
return paths