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| 1 | +// 2583. Kth Largest Sum in a Binary Tree |
| 2 | +// 🟠 Medium |
| 3 | +// |
| 4 | +// https://leetcode.com/problems/kth-largest-sum-in-a-binary-tree/ |
| 5 | +// |
| 6 | +// Tags: Tree - Breadth-First Search - Sorting - Binary Tree |
| 7 | + |
| 8 | +use std::cell::RefCell; |
| 9 | +use std::collections::{BinaryHeap, VecDeque}; |
| 10 | +use std::rc::Rc; |
| 11 | + |
| 12 | +// Definition for a binary tree node. |
| 13 | +#[derive(Debug, PartialEq, Eq)] |
| 14 | +pub struct TreeNode { |
| 15 | + pub val: i32, |
| 16 | + pub left: Option<Rc<RefCell<TreeNode>>>, |
| 17 | + pub right: Option<Rc<RefCell<TreeNode>>>, |
| 18 | +} |
| 19 | + |
| 20 | +impl TreeNode { |
| 21 | + #[inline] |
| 22 | + pub fn new(val: i32) -> Self { |
| 23 | + TreeNode { |
| 24 | + val, |
| 25 | + left: None, |
| 26 | + right: None, |
| 27 | + } |
| 28 | + } |
| 29 | +} |
| 30 | + |
| 31 | +struct Solution; |
| 32 | +impl Solution { |
| 33 | + /// Use BFS to process the tree by level, for each level, compute the sum of it's node |
| 34 | + /// values and store that, then return the kth sum, I use a heap for that, but we could |
| 35 | + /// also use other methods, for example a vector then sort it. |
| 36 | + /// |
| 37 | + /// Time complexity: O(nlog(n)) - We visit each node in the tree in linear time, but we |
| 38 | + /// push/pop them from the heap of size n. |
| 39 | + /// Space complexity: O(n) - The size of the heap. We push one i64 for each level, if the |
| 40 | + /// tree is skeewed, it could have as many levels as nodes. |
| 41 | + /// |
| 42 | + /// Runtime 8 ms Beats 100% |
| 43 | + /// Memory 11.84 MB Beats 33% |
| 44 | + pub fn kth_largest_level_sum(root: Option<Rc<RefCell<TreeNode>>>, k: i32) -> i64 { |
| 45 | + let n = k as usize + 1; |
| 46 | + let mut heap = BinaryHeap::with_capacity(n); |
| 47 | + let mut queue = VecDeque::from([root.unwrap()]); |
| 48 | + let mut level_sum: i64; |
| 49 | + while !queue.is_empty() { |
| 50 | + level_sum = 0; |
| 51 | + for _ in 0..queue.len() { |
| 52 | + if let Some(rc) = queue.pop_front() { |
| 53 | + let node = rc.borrow(); |
| 54 | + level_sum += node.val as i64; |
| 55 | + if let Some(left) = node.left.clone() { |
| 56 | + queue.push_back(left); |
| 57 | + } |
| 58 | + if let Some(right) = node.right.clone() { |
| 59 | + queue.push_back(right); |
| 60 | + } |
| 61 | + } |
| 62 | + } |
| 63 | + heap.push(level_sum); |
| 64 | + } |
| 65 | + for _ in 0..k - 1 { |
| 66 | + heap.pop(); |
| 67 | + } |
| 68 | + *heap.peek().unwrap() |
| 69 | + } |
| 70 | +} |
| 71 | + |
| 72 | +// Tests. |
| 73 | +fn main() { |
| 74 | + let tree1 = Some(Rc::new(RefCell::new(TreeNode { |
| 75 | + val: 5, |
| 76 | + left: Some(Rc::new(RefCell::new(TreeNode { |
| 77 | + val: 8, |
| 78 | + left: Some(Rc::new(RefCell::new(TreeNode { |
| 79 | + val: 2, |
| 80 | + left: Some(Rc::new(RefCell::new(TreeNode { |
| 81 | + val: 4, |
| 82 | + left: None, |
| 83 | + right: None, |
| 84 | + }))), |
| 85 | + right: Some(Rc::new(RefCell::new(TreeNode { |
| 86 | + val: 6, |
| 87 | + left: None, |
| 88 | + right: None, |
| 89 | + }))), |
| 90 | + }))), |
| 91 | + right: Some(Rc::new(RefCell::new(TreeNode { |
| 92 | + val: 1, |
| 93 | + left: None, |
| 94 | + right: None, |
| 95 | + }))), |
| 96 | + }))), |
| 97 | + right: Some(Rc::new(RefCell::new(TreeNode { |
| 98 | + val: 9, |
| 99 | + left: Some(Rc::new(RefCell::new(TreeNode { |
| 100 | + val: 3, |
| 101 | + left: None, |
| 102 | + right: None, |
| 103 | + }))), |
| 104 | + right: Some(Rc::new(RefCell::new(TreeNode { |
| 105 | + val: 7, |
| 106 | + left: None, |
| 107 | + right: None, |
| 108 | + }))), |
| 109 | + }))), |
| 110 | + }))); |
| 111 | + let tree2 = Some(Rc::new(RefCell::new(TreeNode { |
| 112 | + val: 1, |
| 113 | + left: Some(Rc::new(RefCell::new(TreeNode { |
| 114 | + val: 2, |
| 115 | + left: Some(Rc::new(RefCell::new(TreeNode { |
| 116 | + val: 3, |
| 117 | + left: None, |
| 118 | + right: None, |
| 119 | + }))), |
| 120 | + right: None, |
| 121 | + }))), |
| 122 | + right: None, |
| 123 | + }))); |
| 124 | + let tests = [(tree1, 2, 13), (tree2, 1, 3)]; |
| 125 | + println!("\n\x1b[92m» Running {} tests...\x1b[0m", tests.len()); |
| 126 | + let mut success = 0; |
| 127 | + for (i, t) in tests.iter().enumerate() { |
| 128 | + let res = Solution::kth_largest_level_sum(t.0.clone(), t.1); |
| 129 | + if res == t.2 { |
| 130 | + success += 1; |
| 131 | + println!("\x1b[92m✔\x1b[95m Test {} passed!\x1b[0m", i); |
| 132 | + } else { |
| 133 | + println!( |
| 134 | + "\x1b[31mx\x1b[95m Test {} failed expected: {:?} but got {}!!\x1b[0m", |
| 135 | + i, t.1, res |
| 136 | + ); |
| 137 | + } |
| 138 | + } |
| 139 | + println!(); |
| 140 | + if success == tests.len() { |
| 141 | + println!("\x1b[30;42m✔ All tests passed!\x1b[0m") |
| 142 | + } else if success == 0 { |
| 143 | + println!("\x1b[31mx \x1b[41;37mAll tests failed!\x1b[0m") |
| 144 | + } else { |
| 145 | + println!( |
| 146 | + "\x1b[31mx\x1b[95m {} tests failed!\x1b[0m", |
| 147 | + tests.len() - success |
| 148 | + ) |
| 149 | + } |
| 150 | +} |
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