257. Binary Tree Paths LeetCode Solution

In this guide, you will get 257. Binary Tree Paths LeetCode Solution with the best time and space complexity. The solution to Binary Tree Paths problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Binary Tree Paths solution in C++
4. Binary Tree Paths solution in Java
5. Binary Tree Paths solution in Python
6. Additional Resources

Problem Statement of Binary Tree Paths

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

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(h)

257. Binary Tree Paths LeetCode Solution in C++

class Solution {
 public:
  vector<string> binaryTreePaths(TreeNode* root) {
    vector<string> ans;
    dfs(root, {}, ans);
    return ans;
  }

 private:
  void dfs(TreeNode* root, vector<string>&& path, vector<string>& ans) {
    if (root == nullptr)
      return;
    if (root->left == nullptr && root->right == nullptr) {
      ans.push_back(join(path) + to_string(root->val));
      return;
    }

    path.push_back(to_string(root->val) + "->");
    dfs(root->left, std::move(path), ans);
    dfs(root->right, std::move(path), ans);
    path.pop_back();
  }

  string join(const vector<string>& path) {
    string joined;
    for (const string& s : path)
      joined += s;
    return joined;
  }
};
/* code provided by PROGIEZ */

257. Binary Tree Paths LeetCode Solution in Java

class Solution {
  public List<String> binaryTreePaths(TreeNode root) {
    List<String> ans = new ArrayList<>();
    dfs(root, new StringBuilder(), ans);
    return ans;
  }

  private void dfs(TreeNode root, StringBuilder sb, List<String> ans) {
    if (root == null)
      return;
    if (root.left == null && root.right == null) {
      ans.add(sb.append(root.val).toString());
      return;
    }

    final int length = sb.length();
    dfs(root.left, sb.append(root.val).append("->"), ans);
    sb.setLength(length);
    dfs(root.right, sb.append(root.val).append("->"), ans);
    sb.setLength(length);
  }
}
// code provided by PROGIEZ

257. Binary Tree Paths LeetCode Solution in Python

class Solution:
  def binaryTreePaths(self, root: TreeNode | None) -> list[str]:
    ans = []

    def dfs(root: TreeNode | None, path: list[str]) -> None:
      if not root:
        return
      if not root.left and not root.right:
        ans.append(''.join(path) + str(root.val))
        return

      path.append(str(root.val) + '->')
      dfs(root.left, path)
      dfs(root.right, path)
      path.pop()

    dfs(root, [])
    return ans
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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