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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/binary-tree-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 [] children=[root.left,root.right] if not any(children): return ['1'] def dfs(root,path): if root: path+=str(root.val) if not root.left and not root.right: res.append(path) else: path+='->' dfs(root.left,path) dfs(root.right,path) res=[] dfs(root,'') return res
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# 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]: def dfs(root,path): if root: path+=str(root.val) if not root.left and not root.right: res.append(path) else: path+='->' dfs(root.left,path) dfs(root.right,path) res=[] dfs(root,'') return res