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94, Binary Tree In-order Traversal

MikeAbout 2 minbinary treeeasystackbinary treedepth first search

I Problem

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

Example 1
3 nodes
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?

Related Topics

  • Stack
  • Depth-First Search
  • Binary Tree

II Solution

#[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,
        }
    }
}

Approach 1: Recursion

pub fn inorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
    let mut res = vec![];
    const RECURSION_IMPL: fn(root: Option<Rc<RefCell<TreeNode>>>, res: &mut Vec<i32>) =
        |root, res| match root {
            None => {}
            Some(root) => {
                RECURSION_IMPL(root.borrow_mut().left.take(), res);
                res.push(root.borrow().val);
                RECURSION_IMPL(root.borrow_mut().right.take(), res);
            }
        };

    RECURSION_IMPL(root, &mut res);
    return res;
}

Approach 2: Iteration

pub fn inorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
    //Self::iteration_impl_1(root)
    //Self::iteration_impl_2(root)
    Self::iteration_impl_3(root)
}

fn iteration_impl_1(mut root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
    let mut res = vec![];
    let mut stack = vec![];

    while root.is_some() || !stack.is_empty() {
        while let Some(curr) = root {
            root = curr.borrow_mut().left.take();
            stack.push(curr);
        }
        if let Some(curr) = stack.pop() {
            res.push(curr.borrow().val);
            root = curr.borrow_mut().right.take();
        }
    }

    res
}

fn iteration_impl_2(mut root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
    let mut res = vec![];
    let mut stack = vec![];

    while root.is_some() || !stack.is_empty() {
        match root {
            Some(curr) => {
                root = curr.borrow_mut().left.take();
                stack.push(curr);
            }
            None => {
                if let Some(curr) = stack.pop() {
                    res.push(curr.borrow().val);
                    root = curr.borrow_mut().right.take();
                }
            }
        }
    }

    res
}

fn iteration_impl_3(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
    let mut res = vec![];
    
    if let Some(root) = root {
        let mut stack = vec![Ok(root)];
        while let Some(curr) = stack.pop() {
            match curr {
                Ok(node) => {
                    if let Some(right) = node.borrow_mut().right.take() {
                        stack.push(Ok(right));          // Right
                    }
                    stack.push(Err(node.borrow().val)); // Root
                    if let Some(left) = node.borrow_mut().left.take() {
                        stack.push(Ok(left));           // Left
                    }
                }
                Err(val) => res.push(val),
            }
        }
    }

    res
}