Count Complete Tree Nodes

 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.

 Brute Force Approach: DFS (O(n))

Just do any traversal and count nodes.

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


❌ Why not optimal?

  • It's O(n) time → may be too slow for large trees (up to 10⁵ nodes)

  • Doesn’t use the complete tree property

 Optimal Approach: Use Binary Tree Height Trick (O(log²n))

💡 Key Insight:

In a complete binary tree:

  • If left height = right height, it is a perfect binary tree → total nodes = 2^h - 1

  • Otherwise, recursively count in left & right.

🔁 Steps:

  1. Get leftmost height of left subtree

  2. Get rightmost height of right subtree

  3. If heights are equal, it's a perfect subtree → use formula

  4. If not equal → recursively count nodes in left and right

class Solution {
    public int countNodes(TreeNode root) {
        // Base case: If tree is empty, return 0
        if (root == null) return 0;

        // Compute leftmost height (go only left)
        int leftHeight = getLeftHeight(root);

        // Compute rightmost height (go only right)
        int rightHeight = getRightHeight(root);

        // If both heights are equal, it is a perfect binary tree
        if (leftHeight == rightHeight) {
            // Total nodes in a perfect binary tree = 2^h - 1
            return (1 << leftHeight) - 1;
        }

        // Otherwise, count recursively: 1 for root + left subtree + right subtree
        return 1 + countNodes(root.left) + countNodes(root.right);
    }

    // Helper function to calculate height by going left
    private int getLeftHeight(TreeNode node) {
        int height = 0;
        while (node != null) {
            height++;           // Increment for each level
            node = node.left;   // Move left
        }
        return height;
    }

    // Helper function to calculate height by going right
    private int getRightHeight(TreeNode node) {
        int height = 0;
        while (node != null) {
            height++;           // Increment for each level
            node = node.right;  // Move right
        }
        return height;
    }
}

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