222. Count Complete Tree Nodes

Problem

Difficulty: Easy
Tags: Binary Search, Bit Manipulation, Tree, Binary Tree

Intuition

It seemed there is beutiful way to count nodes only for complete node. but I couldn’t find it. so I got little help from internet.

Approach

the apprach is simple. it is just like counting algorithm for any tree, but it has some additional feature that returns $2^h -1$ if the tree is fully filled.

Solution

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */


int countNodes(struct TreeNode* root) {
    if(!root){
        return 0;
    }
    struct TreeNode* tmpL = root;
    struct TreeNode* tmpR = root;

    int hL = 1;
    int hR = 1;
    while(tmpL->left){
        tmpL = tmpL->left;
        hL += 1;
    }
    while(tmpR->right){
        tmpR = tmpR->right;
        hR += 1;
    }

    if (hL == hR){
        return pow(2, hL) -1;
    }

    return 1 + countNodes(root->right) + countNodes(root->left);
}

Complexity

Time: $O(\ln(n)^2)$
Space: $O(\ln(n))$

Thoughts

this time I overfocused on the fact of “complete tree”. I need to focus more on the solution. not the problem and restrictions.