# 222. Count Complete Tree Nodes

## Description

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.

Design an algorithm that runs in less than O(n) time complexity.

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.

## Solutions

Solution 1: Recursion

We recursively traverse the entire tree and count the number of nodes.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of nodes in the tree.

Solution 2: Binary Search

For this problem, we can also take advantage of the characteristics of a complete binary tree to design a faster algorithm.

Characteristics of a complete binary tree: leaf nodes can only appear on the bottom and second-to-bottom layers, and the leaf nodes on the bottom layer are concentrated on the left side of the tree. It should be noted that a full binary tree is definitely a complete binary tree, but a complete binary tree is not necessarily a full binary tree.

If the number of layers in a full binary tree is $h$, then the total number of nodes is $2^h - 1$.

We first count the heights of the left and right subtrees of $root$, denoted as $left$ and $right$.

1. If $left = right$, it means that the left subtree is a full binary tree, so the total number of nodes in the left subtree is $2^{left} - 1$. Plus the $root$ node, it is $2^{left}$. Then we recursively count the right subtree.
2. If $left > right$, it means that the right subtree is a full binary tree, so the total number of nodes in the right subtree is $2^{right} - 1$. Plus the $root$ node, it is $2^{right}$. Then we recursively count the left subtree.

The time complexity is $O(\log^2 n)$. (square of tree-height)

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public int countNodes(TreeNode root) {
if (root == null) {
return 0;
}
int left = depth(root.left);
int right = depth(root.right);
if (left == right) {
return (1 << left) + countNodes(root.right);
}
return (1 << right) + countNodes(root.left);
}

private int depth(TreeNode root) {
int d = 0;
for (; root != null; root = root.left) {
++d;
}
return d;
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int countNodes(TreeNode* root) {
if (!root) {
return 0;
}
int left = depth(root->left);
int right = depth(root->right);
if (left == right) {
return (1 << left) + countNodes(root->right);
}
return (1 << right) + countNodes(root->left);
}

int depth(TreeNode* root) {
int d = 0;
for (; root; root = root->left) {
++d;
}
return d;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def countNodes(self, root: Optional[TreeNode]) -> int:
def depth(root):
d = 0
while root:
d += 1
root = root.left
return d

if root is None:
return 0
left, right = depth(root.left), depth(root.right)
if left == right:
# left child subtree: (1<<left)-1
# plus root: +1
# so total except right subtree: (1<<left)
return (1 << left) + self.countNodes(root.right)
else: # left = right+1
return (1 << right) + self.countNodes(root.left)

############

'''
>>> 2 ** 3
8
>>> 3 ** 2
9
'''

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
def getHeight(self, root):
height = 0
while root:
height += 1
root = root.left
return height

def countNodes(self, root):
count = 0
while root:
l, r = map(self.getHeight, (root.left, root.right))
if l == r:
count += 2 ** l
root = root.right
else:
count += 2 ** r
root = root.left
return count


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func countNodes(root *TreeNode) int {
if root == nil {
return 0
}
left, right := depth(root.Left), depth(root.Right)
if left == right {
return (1 << left) + countNodes(root.Right)
}
return (1 << right) + countNodes(root.Left)
}

func depth(root *TreeNode) (d int) {
for ; root != nil; root = root.Left {
d++
}
return
}

• /**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var countNodes = function (root) {
const depth = root => {
let d = 0;
for (; root; root = root.left) {
++d;
}
return d;
};
if (!root) {
return 0;
}
const left = depth(root.left);
const right = depth(root.right);
if (left == right) {
return (1 << left) + countNodes(root.right);
}
return (1 << right) + countNodes(root.left);
};


• /**
* Definition for a binary tree node.
* public class TreeNode {
*     public int val;
*     public TreeNode left;
*     public TreeNode right;
*     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
public class Solution {
public int CountNodes(TreeNode root) {
if (root == null) {
return 0;
}
int left = depth(root.left);
int right = depth(root.right);
if (left == right) {
return (1 << left) + CountNodes(root.right);
}
return (1 << right) + CountNodes(root.left);
}

private int depth(TreeNode root) {
int d = 0;
for (; root != null; root = root.left) {
++d;
}
return d;
}
}

• use std::cell::RefCell;
use std::rc::Rc;

impl Solution {
pub fn count_nodes(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
if let Some(node) = root {
let node = node.borrow();
let left = Self::depth(&node.left);
let right = Self::depth(&node.right);
if left == right {
Self::count_nodes(node.right.clone()) + (1 << left)
} else {
Self::count_nodes(node.left.clone()) + (1 << right)
}
} else {
0
}
}

fn depth(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
if let Some(node) = root { Self::depth(&node.borrow().left) + 1 } else { 0 }
}
}