0222. 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.

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.

Follow up: Traversing the tree to count the number of nodes in the tree is an easy solution but with O(n) complexity. Could you find a faster algorithm?

Analysis

A complete binary tree has the special property: at the last level of the binary tree, the nodes will be populated from left to right. There could be empty/null nodes at any point in the last level, but from the first empty/null node, all the rest to the right have to be empty/null as well.

If we are given a perfect binary tree, we can use $2^{height} - 1$ to calculate the total number of nodes in the tree. The problem now breaks down to finding the position starting at which the rest of the leaf nodes become empty/null. We can use binary search to find the pivot node.

1. start from left and right, and traverse $\log(n)$ levels to see the heights of left and right.
2. if left and right heights are the same, it means we are already in a perfect binary tree -> we can use the formula to calculate the number of nodes in the current subtree.

3. if left = right + 1, then the pivot should be somewhere between left and right, so we should recursively call both.

4. Time: $\log^2 (n)$

5. Space: $\log(n)$ -> it takes $\log(n)$ times to find the pivot, thus it will call the recursion $\log(n)$ times.

Code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int countNodes(TreeNode* root) {
        if (!root) return 0;
        int l = 0, r = 0;
        TreeNode *a = root, *b = root;
        while (a) {
            a = a -> left;
            l++;
        }
        while (b) {
            b = b -> right;
            r++;
        }
        if (l == r) return pow(2, l) - 1;
        return countNodes(root -> left) + countNodes(root -> right) + 1;
    }
};