222. 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.

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:-

Using Recursion :-

	 class Solution {
	public:
		int countNodes(TreeNode* root) {
			if(root==NULL)
			{
				return 0;
			}
			else
			{
				return countNodes(root->left)+countNodes(root->right)+1;
			}
		}
	};

Time Complexity :- O(N)
Space Complexity :- O(N)

Optimal Solution :-

class Solution {
	public:

	int countNodes(TreeNode* root) {
		int lh=0,rh=0;
		TreeNode* curr=root;
		while(curr!= NULL)
		{
			lh++;
			curr=curr->left;
		}
	   curr=root;
    
    while(curr!= NULL)
    {
        rh++;
        curr=curr->right;
    }
    if(lh==rh)
    {
        return pow(2,lh)-1;
    }
    else
    {
        return countNodes(root->left)+countNodes(root->right)+1;
    }
 }
};

Time Complexity :- O( (logN)^2 )
Space Complexity :- O(1)

Properties :-
A complete binary tree has (2^n – 1) nodes in total.