Find bottom left tree value

Tree Depth-First Search Breadth-First Search Binary Tree

72.2% acceptance

Feb 19, 2026

3993

300

Given the root of a binary tree, return the leftmost value in the last row of the tree.

Solution

Rust

Time O(n²)

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;
use std::collections::VecDeque;
impl Solution {
  pub fn find_bottom_left_value(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
    let mut result = 0;
    let mut queue = VecDeque::new();
    if let Some(r) = root { queue.push_back(r); }
    while !queue.is_empty() {
      let size = queue.len();
      for i in 0..size {
        let node = queue.pop_front().unwrap();
        let nb = node.borrow();
        if i == 0 { result = nb.val; }
        if let Some(l) = nb.left.clone() { queue.push_back(l); }
        if let Some(r) = nb.right.clone() { queue.push_back(r); }
      }
    }
    result
  }
}