222. Count Complete Tree Nodes

Given a complete binary tree, count the number of nodes.

Definition of a complete binary tree fromWikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2hnodes inclusive at the last level h.

Analysis:

The height of a tree can be found by just going left. Let a single node tree have height 0. Find the heighthof the whole tree. If the whole tree is empty, i.e., has height -1, there are 0 nodes.

Otherwise check whether the height of the right subtree is just one less than that of the whole tree, meaning left and right subtree have the same height.

• If yes, then the last node on the last tree row is in the right subtree and the left subtree is a full tree of height h-1. So we take the 2^h-1 nodes of the left subtree plus the 1 root node plus recursively the number of nodes in the right subtree.
• If no, then the last node on the last tree row is in the left subtree and the right subtree is a full tree of height h-2. So we take the 2^(h-1)-1 nodes of the right subtree plus the 1 root node plus recursively the number of nodes in the left subtree.

Since I halve the tree in every recursive step, I have O(log(n)) steps. Finding a height costs O(log(n)). So overall O(log(n)^2).

Complexity:

Time: O(log(n)^2)

Space: O(logn)

Code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int height(TreeNode root) {
        return root == null ? -1 : 1 + height(root.left);
    }
    public int countNodes(TreeNode root) {
        int h = height(root);
        return h < 0 ? 0 :
               height(root.right) == h-1 ? (1 << h) + countNodes(root.right)
                                         : (1 << h-1) + countNodes(root.left);
    }
}