README_EN.md

257. Binary Tree Paths

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Description

Given the root of a binary tree, return all root-to-leaf paths in any order.

A leaf is a node with no children.

 

Example 1:

Input: root = [1,2,3,null,5]
Output: ["1->2->5","1->3"]

Example 2:

Input: root = [1]
Output: ["1"]

 

Constraints:

• The number of nodes in the tree is in the range [1, 100].
• -100 <= Node.val <= 100

Solutions

Python3

# 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):
            if root is None:
                return
            path.append(str(root.val))
            if root.left is None and root.right is None:
                res.append("->".join(path))
            dfs(root.left)
            dfs(root.right)
            path.pop()
        res = []
        path = []
        dfs(root)
        return res

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private List<String> res;
    private List<String> path;

    public List<String> binaryTreePaths(TreeNode root) {
        if (root == null) return Collections.emptyList();
        res = new ArrayList<>();
        path = new ArrayList<>();
        dfs(root);
        return res;
    }

    private void dfs(TreeNode root) {
        if (root == null) return;
        path.add(String.valueOf(root.val));
        if (root.left == null && root.right == null) {
            res.add(String.join("->", path));
        }
        dfs(root.left);
        dfs(root.right);
        path.remove(path.size() - 1);
    }
}

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