Binary tree paths

String Backtracking Tree Depth-First Search Binary Tree

68.2% acceptance

Feb 27, 2026

7210

346

Given the root of a binary tree, return all root-to-leaf paths in any order. A leaf is a node with no children.

Solution

Rust

Time O(n)

Space O(n)

LeetCode

solution.rs

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
// 
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
  pub fn binary_tree_paths(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<String> {
    let mut result = Vec::new();
    if let Some(node) = root {
      Self::tree_dfs(&node, String::new(), &mut result);
    }
    result
  }
  
  fn tree_dfs(node: &Rc<RefCell<TreeNode>>, path: String, result: &mut Vec<String>) {
    let node = node.borrow();
    let new_path = if path.is_empty() {
      node.val.to_string()
    } else {
      format!("{}->{}",path, node.val)
    };
    
    if node.left.is_none() && node.right.is_none() {
      result.push(new_path);
      return;
    }
    
    if let Some(left) = &node.left {
      Self::tree_dfs(left, new_path.clone(), result);
    }
    if let Some(right) = &node.right {
      Self::tree_dfs(right, new_path, result);
    }
  }
}