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algorithm-review/chapter10/3_binary-tree-paths.py
T
2020-06-22 10:08:57 +08:00

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Python

#!/usr/bin/env python
# coding=utf-8
#######################################################################################
# Leetcode 257 二叉树的所有路径
#
# 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
#######################################################################################
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
def __repr__(self):
if self:
return "{}->{}->{}".format(self.val, repr(self.left), repr(self.right))
class Solution:
def binaryTreePaths(self, root):
"""
:type root: TreeNode
:rtype List[str]
(knowledge)
思路:
1. 递归的进行先序遍历,过程中记录走过的路径;
2. 当遍历到一个叶子节点时,将当前路径添加到结果集当中;
"""
def preorderTraversal(root, s, result):
if not root:
return
if not root.left and not root.right:
result.append(s + "->" + str(root.val))
preOrderTraversal(root.left, s + "->" + str(root.val), result)
preOrderTraversal(root.right, s + "->" + str(root.val), result)
if not root:
return []
if not root.left and not root.right:
return [str(root.val)]
s = str(root.val)
result = []
preOrderTraversal(root, s, result)
return result