Binary tree inorder traversal

Stack Tree Depth-First Search Binary Tree

79.8% acceptance

Feb 27, 2026

14828

895

Given the root of a binary tree, return the inorder traversal of its nodes' values.

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 inorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
    let mut result = Vec::new();
    Self::inorder(root.as_ref(), &mut result);
    result
  }
  
  fn inorder(node: Option<&Rc<RefCell<TreeNode>>>, result: &mut Vec<i32>) {
    if let Some(n) = node {
      let n = n.borrow();
      Self::inorder(n.left.as_ref(), result);
      result.push(n.val);
      Self::inorder(n.right.as_ref(), result);
    }
  }
}