Binary Tree Path

Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

   1
 /   \
2     3
 \
  5

All root-to-leaf paths are:

["1->2->5", "1->3"]

Solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param {TreeNode} root
    # @return {string[]}
    def binaryTreePaths(self, root):
        if root is None:
            return []
        result, tmp = [], []
        self.dfs(root, result, tmp)
        return result

    def dfs(self, root, result, tmp):
        if root.right is None and root.left is None:
            tmp.append(root.val)
            result.append("->".join(str(x) for x in tmp))
            tmp.pop()
        else:
            tmp.append(root.val)
            if root.left:
                self.dfs(root.left, result, tmp)
            if root.right:
                self.dfs(root.right, result, tmp)
            tmp.pop()

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

        helper(root, result, new ArrayList<String>());

        return result;
    }

    public void helper(TreeNode root, List<String> result, List<String> path) {
        path.add(Integer.toString(root.val));
        if(root.left == null && root.right == null) {
            result.add(String.join("->", path));
        } else {
            if(root.left != null)
                helper(root.left, result, path);
            if(root.right != null)
                helper(root.right, result, path);
        }
        path.remove(path.size() - 1);
    }
}