Count Complete Tree Nodes

Count Complete Tree Nodes - Problem

Given the root of a complete binary tree, return the number of 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^h nodes inclusive at the last level h.

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

Input & Output

Example 1 — Complete Binary Tree

$ Input: root = [1,2,3,4,5,6]

› Output: 6

💡 Note: All levels are filled except the last level has 3 out of 4 possible nodes, total count is 6

Example 2 — Single Node

$ Input: root = [1]

› Output: 1

💡 Note: Tree has only root node, so count is 1

Example 3 — Perfect Binary Tree

$ Input: root = [1,2,3,4,5,6,7]

› Output: 7

💡 Note: All levels are completely filled, total 7 nodes in perfect binary tree

Constraints

The number of nodes in the tree is in the range [1, 2 × 10⁴]
1 ≤ Node.val ≤ 5 × 10⁴
The tree is guaranteed to be complete

Visualization

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Asked in

G Google 12 f Facebook 8 a Amazon 6 M Microsoft 5

The key insight is leveraging the complete binary tree property to avoid visiting every node. The optimal approach uses binary search on the last level combined with path existence checking to achieve O(log² n) time complexity instead of the naive O(n). Best approach: Binary Search - Time: O(log² n), Space: O(log n).

Common Approaches

✓ Binary Search on Last Level

⏱️ Time: O(log² n) Space: O(log n)

Calculate the height of the tree, then use binary search to find how many nodes exist in the last level. For a complete binary tree, we can determine if a node exists by checking if the path to it is valid.

Brute Force DFS

⏱️ Time: O(n) Space: O(h)

Traverse the entire tree using DFS (or BFS) and count each node visited. This approach doesn't take advantage of the complete binary tree property.

Binary Search on Last Level — Algorithm Steps

Step 1: Calculate the height of the tree
Step 2: Check if tree is perfect (last level full)
Step 3: Binary search on last level positions
Step 4: Return total count using formula

Visualization

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Step-by-Step Walkthrough

Calculate Height

Find tree height by going left

Binary Search Range

Search positions 0 to 2^h-1 in last level

Check Existence

Use path following to verify if node exists

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
};

int solution(struct TreeNode* root) {
    if (!root) return 0;
    return 1 + solution(root->left) + solution(root->right);
}

struct TreeNode* newNode(int val) {
    struct TreeNode* node = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    node->val = val;
    node->left = NULL;
    node->right = NULL;
    return node;
}

int main() {
    struct TreeNode* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    
    int result = solution(root);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log² n)

Height calculation O(log n) + binary search O(log n) with O(log n) existence check

⚡ Linearithmic

Space Complexity

O(log n)

Recursion stack for height calculation and path checking

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Binary Search on Last Level — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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