94. Binary Tree Inorder Traversal

Description of Problem

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

Example 1:

Input: root = [1,null,2,3]
Output: [1,3,2]

Example 2:

Input: root = []
Output: []

Example 3:

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

Constraints:

The number of nodes in the tree is in the range [0, 100].
100 <= Node.val <= 100

Follow up: Recursive solution is trivial, could you do it iteratively?

Solution (Failed to answer the follow-up)

Tags: Binary Tree Rust

Code (Rust) - Recursion

// 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::borrow::BorrowMut;
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    pub fn inorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut v : Vec<i32> = Vec::new();
        Solution::helper(root, &mut v);
        return v;

    }

    fn helper(node: Option<Rc<RefCell<TreeNode>>>, v : &mut Vec<i32>){
        if let Some(node) = node{
            let node = node.borrow();
            Solution::helper(node.left.clone(), v);
            v.push(node.val);
            Solution::helper(node.right.clone(), v);
        }
    }
}

Complexity

n is the number of elements in the tree
h is the height of the tree

Time Complexity

\( T(n) = O(n) \)
- Traverse all nodes in tree

Auxiliary Space

\( S(h) = O(h) \)

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