# Question

Formatted question description: https://leetcode.ca/all/222.html

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.

# Algorithm

First, we need a getHeight function, which is used to count the maximum height of the left subtree of the current node.

• Call the getHeight function on the current node to get the maximum height h of the left subtree. If returned is -1, it means that the current node does not exist and returns 0 directly.
• Otherwise, call the getHeight function on the right child.
• If the return value is h-1, it means that the left sub-tree is a perfect binary tree, and the number of nodes in the left sub-tree is 2^h-1. Adding the current node, there are 2^h in total, that is, 1<<h,
• then, add the return value of calling the recursive function on the right child node.
• If the return value of calling the getHeight function on the right subnode is not h-1, it means that the right subtree must be a perfect tree, and the height is h-1,
• then the number of summary points is 2^(h-1) -1, plus the current node is 2^(h-1), that is, 1<<(h-1), and then add the return value of calling the recursive function on the left child node

# Code

• 
public class Count_Complete_Tree_Nodes {
/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/

public class Solution_optimize {

public int countNodes(TreeNode root) {

int res = 0;
int h = getHeight(root); // here h is actually h-1

if (h < 0) return 0;

if (getHeight(root.right) == h - 1) {
return (1 << h) + countNodes(root.right);
} else { // getHeight(root.right) == h - 2
return (1 << (h - 1)) + countNodes(root.left);
}
}

int getHeight(TreeNode node) {
return node != null ? (1 + getHeight(node.left)) : -1;
}
}

public class Solution {
public int countNodes(TreeNode root) {
if (root == null) {
return 0;
}

int leftHeight = findLeftHeight(root);
int rightHeight = findRightHeight(root);

if (leftHeight == rightHeight) {
return (2 << (leftHeight - 1)) - 1;
}

return 1 + countNodes(root.left) + countNodes(root.right);
}

private int findLeftHeight(TreeNode root) {
if (root == null) {
return 0;
}

int height = 1;

while (root.left != null) {
height++;
root = root.left;
}

return height;
}

private int findRightHeight(TreeNode root) {
if (root == null) {
return 0;
}

int height = 1;

while (root.right != null) {
height++;
root = root.right;
}

return 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;
}
}

• // OJ: https://leetcode.com/problems/count-complete-tree-nodes/
// Time: O(H^2)
// Space: O(H)
class Solution {
int countLeft(TreeNode *root) {
int cnt = 0;
for (; root; ++cnt, root = root->left);
return cnt;
}
int countRight(TreeNode *root) {
int cnt = 0;
for (; root; ++cnt, root = root->right);
return cnt;
}
public:
int countNodes(TreeNode* root) {
if (!root) return 0;
int left = countLeft(root), right = countRight(root);
if (left == right) return (1 << left) - 1;
return countNodes(root->left) + countNodes(root->right) + 1;
}
};

• # 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
}
}
}