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 `2`

nodes inclusive at the last level ^{h}`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 * 10`

.^{4}] `0 <= Node.val <= 5 * 10`

^{4}- The tree is guaranteed to be
**complete**.

`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);
}
```