Count Complete Tree Nodes
Problem Description
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:
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.
Solve it on LeetCode
Approaches
1. Simple Tree Traversal
Intuition:
The most straightforward method to count all the nodes in a tree is to traverse the tree and count each node. In a normal binary tree, this approach works perfectly fine.
Steps:
- Use Depth-First Search (DFS) to traverse each node starting from the root.
- For every node visited, increment a count.
- The total count after the complete traversal will give the number of nodes in the tree.
Code:
- Time Complexity: O(n), where n is the number of nodes in the tree. We visit each node exactly once.
- Space Complexity: O(h), where h is the height of the tree. This is due to the recursion stack.
2. Binary Search and Depth Calculation
Intuition:
Given the properties of a complete tree, we can use a more efficient method leveraging the structure. The idea is to take advantage of the depth of the tree and binary search to determine missing nodes in the last level.
Steps:
- Calculate the depth (measured by the number of edges) of the tree using the left-most path.
- If the tree is empty, return 0.
- Otherwise, perform binary search on the last level to check if nodes exist:
- Half of the potential nodes can be skipped at each level of the last row, providing savings.
- Calculate the total number of nodes using full levels and the nodes you've confirmed in the last level.
Code:
- Time Complexity: O(log^2(n)), where n is the number of nodes. We perform a log(n) depth computation and an additional log(n) binary search for each depth level.
- Space Complexity: O(1), as it uses constant extra space.