222. Count Complete Tree Nodes #

Problem Statement: #

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.

Solution: #

// Approach 1
// Time Complexity -> O(n)
public int countNodes(TreeNode root) {
    if (root == null)
        return 0;
    return countNodes(root.left) + countNodes(root.right) + 1;
}

// Approach 2
// Time Complexity -> O(logn * logn)
public int countNodes2(TreeNode root) {
    int leftHeight = 0, rightHeight = 0;
    TreeNode currentNode = root;
    while (currentNode != null) {
        leftHeight++;
        currentNode = currentNode.left;
    }
    currentNode = root;
    while (currentNode != null) {
        rightHeight++;
        currentNode = currentNode.right;
    }
    if (leftHeight == rightHeight)
        return (int) Math.pow(2, leftHeight) - 1;
    return countNodes2(root.left) + countNodes2(root.right) + 1;
}

... #