#145
Easy Algorithms Binary tree postorder traversal
Stack Tree Depth-First Search Binary Tree
77.7% acceptance
Feb 27, 2026
7756
223
Given the root of a binary tree, return the postorder traversal of its nodes' values.
Solution
Rust
Time O(n²)
Space O(n)
// 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 postorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
let mut result = Vec::new();
let mut stack = Vec::new();
let mut last_visited: Option<Rc<RefCell<TreeNode>>> = None;
let mut current = root;
while current.is_some() || !stack.is_empty() {
while let Some(node) = current {
stack.push(node.clone());
current = node.borrow().left.clone();
}
if let Some(peek) = stack.last() {
let peek_ref = peek.borrow();
if peek_ref.right.is_some()
&& !last_visited.as_ref().map_or(false, |lv| Rc::ptr_eq(lv, peek_ref.right.as_ref().unwrap())) {
current = peek_ref.right.clone();
} else {
drop(peek_ref);
let node = stack.pop().unwrap();
result.push(node.borrow().val);
last_visited = Some(node);
}
}
}
result
}
}