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## Problem

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


## Code

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public List<String> binaryTreePaths(TreeNode root) {
List<String> res = new ArrayList<>();
if(root == null) return res;

helper(res, root, "");
return res;
}

private void helper(List<String> res, TreeNode root, String curr) {
if(root == null) return;
if(root.left == null && root.right == null) {
return;
}

helper(res, root.left, curr + root.val + "->");
helper(res, root.right, curr + root.val + "->");
}
}

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def helper(self, res: List[str], root: TreeNode, curr: str):
if not root:
return
if not root.left and not root.right:
res.append(curr + str(root.val))
return

if root.left:
self.helper(res, root.left, curr + str(root.val) + "->")

if root.right:
self.helper(res, root.right, curr + str(root.val) + "->")

def binaryTreePaths(self, root: TreeNode) -> List[str]:
res = []
self.helper(res, root, "")
return res