# Count Nodes in Complete Binary Tree - Google Top Interview Questions

### Problem Statement :

Given a complete binary tree root, return the number of nodes in the tree.

This should be done in \mathcal{O}((\log n)^2)O((logn)
2
).

Constraints

n ≤ 100,000 where n is the number of nodes in root

Example 1

Input

root = [1, [2, [4, null, null], [5, null, null]], [3, null, null]]

Output

5

Example 2

Input

root = [1, [2, [4, null, null], [5, null, null]], [3, [6, null, null], [7, null, null]]]

Output

7

### Solution :

                        Solution in C++ :

int solve(Tree* tree) {
int right_h = 0, left_h = 0;
auto* curr = tree;
while (curr) right_h++, curr = curr->right;
curr = tree;
while (curr) left_h++, curr = curr->left;
if (right_h ==
left_h) {  // if left_height and right_height is same, then the tree has (2**h - 1) nodes
return (1 << right_h) - 1;
}
// If not same, then again make a recursive call on the left and right subtree
return solve(tree->left) + solve(tree->right) + 1;
}


                        Solution in Java :

import java.util.*;

/**
* public class Tree {
*   int val;
*   Tree left;
*   Tree right;
* }
*/
class Solution {
public int solve(Tree tree) {
int lo = 1;
int hi = 100000;
while (lo < hi) {
int m = (lo + hi + 1) / 2;
boolean exists = check(tree, m);
if (exists)
lo = m;
else
hi = m - 1;
}
return lo;
}

public boolean check(Tree t, int m) {
boolean active = false;
for (int i = 17; i >= 0; i--) {
int cur = m & (1 << i);
if (!active) {
if (cur > 0)
active = true;
} else {
if (cur == 0)
t = t.left;
else
t = t.right;
if (t == null)
return false;
}
}
return true;
}
}


                        Solution in Python :

class Solution:
def solve(self, tree):

# function to find the left most depth or  the right most depth
def extreme(root, left):
height = 1
if left:
while root:
root = root.left
height += 1
else:
while root:
root = root.right
height += 1
return height

# main function to solve the problem

def traverse(root):
if not root:
return 0
l = extreme(root.left, True)
r = extreme(root.right, False)
if l == r:  # encountered a full binary tree
return 2 ** l - 1
else:
return traverse(root.left) + traverse(root.right) + 1

ans = traverse(tree)
return ans


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