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
算法一:BFS
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
def binaryTreePaths(self, root: TreeNode):
from collections import deque
if not root: return []
res = []
queue = deque()
queue.appendleft([root, []])
while queue:
node, tmp = queue.pop()###最重要的就是这个temp,当当前节点左存在的时候,把当前节点的值记录下来。这样一直记录就能够记录到叶子节点。叶子节点左右为空。返回res
if not node.left and not node.right:
res.append("->".join(tmp + [str(node.val)]))
if node.left:
queue.appendleft([node.left, tmp + [str(node.val)]])
if node.right:
queue.appendleft([node.right, tmp + [str(node.val)]])
return res
root = TreeNode(1)
a = TreeNode(2)
b = TreeNode(3)
c = TreeNode(5)
root.left = a
root.right = b
a.right = c
print(Solution().binaryTreePaths(root))
算法二:DFS
class Solution:
def binaryTreePaths(self, root: TreeNode) -> List[str]:
if not root: return []
res = []
def helper(root, tmp):
if not root.left and not root.right:res.append(tmp + [str(root.val)])
if root.left:helper(root.left, tmp + [str(root.val)])
if root.right:helper(root.right, tmp + [str(root.val)])
helper(root, [])
return ["->".join(a) for a in res]