Given the root of a complete binary tree, return the number of the nodes in the tree.
According to Wikipedia, every level, except possibly the last, is completely filled in a complete binary tree, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.
Example 1:
Input: root = [1,2,3,4,5,6] Output: 6
Example 2:
Input: root = [] Output: 0
Example 3:
Input: root = [1] Output: 1
Constraints:
- The number of nodes in the tree is in the range
[0, 5 * 104]. 0 <= Node.val <= 5 * 104- The tree is guaranteed to be complete.
Follow up: Traversing the tree to count the number of nodes in the tree is an easy solution but with
O(n) complexity. Could you find a faster algorithm?struct Solution;
use rustgym_util::*;
trait Count {
fn count(&self) -> i32;
}
impl Count for TreeLink {
fn count(&self) -> i32 {
if let Some(node) = self {
1 + node.borrow().left.count() + node.borrow().right.count()
} else {
0
}
}
}
impl Solution {
fn count_nodes(root: TreeLink) -> i32 {
root.count()
}
}
#[test]
fn test() {
let root = tree!(1, tree!(2, tree!(4), tree!(5)), tree!(3, tree!(6), None));
let res = 6;
assert_eq!(Solution::count_nodes(root), res);
}