RUST GYM Leetcode 662 Maximum Width of Binary Tree Rust Solution

662. Maximum Width of Binary Tree

Given a binary tree, write a function to get the maximum width of the given tree. The maximum width of a tree is the maximum width among all levels.

The width of one level is defined as the length between the end-nodes (the leftmost and right most non-null nodes in the level, where the null nodes between the end-nodes are also counted into the length calculation.

It is guaranteed that the answer will in the range of 32-bit signed integer.

Example 1:

Input: 

           1
         /   \
        3     2
       / \     \  
      5   3     9 

Output: 4
Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).

Example 2:

Input: 

          1
         /  
        3    
       / \       
      5   3     

Output: 2
Explanation: The maximum width existing in the third level with the length 2 (5,3).

Example 3:

Input: 

          1
         / \
        3   2 
       /        
      5      

Output: 2
Explanation: The maximum width existing in the second level with the length 2 (3,2).

Example 4:

Input: 

          1
         / \
        3   2
       /     \  
      5       9 
     /         \
    6           7
Output: 8
Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).

Constraints:

The given binary tree will have between 1 and 3000 nodes.

Rust Solution

struct Solution;
use rustgym_util::*;
use std::collections::HashMap;

trait Preorder {
    fn preorder(
        &self,
        row: usize,
        pos: u32,
        min: &mut HashMap<usize, u32>,
        max: &mut HashMap<usize, u32>,
        diff: &mut u32,
    );
}

impl Preorder for TreeLink {
    fn preorder(
        &self,
        row: usize,
        pos: u32,
        min: &mut HashMap<usize, u32>,
        max: &mut HashMap<usize, u32>,
        diff: &mut u32,
    ) {
        if let Some(node) = self {
            min.entry(row).or_insert(pos);
            max.entry(row).or_insert(pos);
            *min.get_mut(&row).unwrap() = min[&row].min(pos);
            *max.get_mut(&row).unwrap() = max[&row].max(pos);
            *diff = (*diff).max(max[&row] - min[&row] + 1);
            let node = node.borrow();
            node.left.preorder(row + 1, pos << 1, min, max, diff);
            node.right.preorder(row + 1, (pos << 1) + 1, min, max, diff);
        }
    }
}

impl Solution {
    fn width_of_binary_tree(root: TreeLink) -> i32 {
        let mut min: HashMap<usize, u32> = HashMap::new();
        let mut max: HashMap<usize, u32> = HashMap::new();
        let mut res = 0;
        root.preorder(0, 0, &mut min, &mut max, &mut res);
        res as i32
    }
}

#[test]
fn test() {
    let root = tree!(1, tree!(3, tree!(5), tree!(3)), tree!(2, None, tree!(9)));
    let res = 4;
    assert_eq!(Solution::width_of_binary_tree(root), res);
    let root = tree!(1, tree!(3, tree!(5), tree!(3)), None);
    let res = 2;
    assert_eq!(Solution::width_of_binary_tree(root), res);
    let root = tree!(1, tree!(3, tree!(5), None), tree!(2));
    let res = 2;
    assert_eq!(Solution::width_of_binary_tree(root), res);
    let root = tree!(
        1,
        tree!(3, tree!(5, tree!(6), None), None),
        tree!(2, None, tree!(9, None, tree!(7)))
    );
    let res = 8;
    assert_eq!(Solution::width_of_binary_tree(root), res);
}

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