# Question

Formatted question description: https://leetcode.ca/all/222.html

 222	Count Complete Tree Nodes

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

Note:
Definition of a complete binary tree from Wikipedia:
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 2^h nodes inclusive at the last level h.

Example:

Input:
1
/ \
2   3
/ \  /
4  5 6

Output: 6

@tag-tree


# Algorithm

First, we need a getHeight function, which is used to count the maximum height of the left subtree of the current node.

• Call the getHeight function on the current node to get the maximum height h of the left subtree. If returned is -1, it means that the current node does not exist and returns 0 directly.
• Otherwise, call the getHeight function on the right child.
• If the return value is h-1, it means that the left sub-tree is a perfect binary tree, and the number of nodes in the left sub-tree is 2^h-1. Adding the current node, there are 2^h in total, that is, 1<<h,
• then, add the return value of calling the recursive function on the right child node.
• If the return value of calling the getHeight function on the right subnode is not h-1, it means that the right subtree must be a perfect tree, and the height is h-1,
• then the number of summary points is 2^(h-1) -1, plus the current node is 2^(h-1), that is, 1<<(h-1), and then add the return value of calling the recursive function on the left child node

# Code

Java

• 
public class Count_Complete_Tree_Nodes {
/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/

public class Solution_optimize {

public int countNodes(TreeNode root) {

int res = 0;
int h = getHeight(root); // here h is actually h-1

if (h < 0) return 0;

if (getHeight(root.right) == h - 1) {
return (1 << h) + countNodes(root.right);
} else { // getHeight(root.right) == h - 2
return (1 << (h - 1)) + countNodes(root.left);
}
}

int getHeight(TreeNode node) {
return node != null ? (1 + getHeight(node.left)) : -1;
}
}

public class Solution {
public int countNodes(TreeNode root) {
if (root == null) {
return 0;
}

int leftHeight = findLeftHeight(root);
int rightHeight = findRightHeight(root);

if (leftHeight == rightHeight) {
return (2 << (leftHeight - 1)) - 1;
}

return 1 + countNodes(root.left) + countNodes(root.right);
}

private int findLeftHeight(TreeNode root) {
if (root == null) {
return 0;
}

int height = 1;

while (root.left != null) {
height++;
root = root.left;
}

return height;
}

private int findRightHeight(TreeNode root) {
if (root == null) {
return 0;
}

int height = 1;

while (root.right != null) {
height++;
root = root.right;
}

return height;
}
}
}

• // OJ: https://leetcode.com/problems/count-complete-tree-nodes/
// Time: O(H^2)
// Space: O(H)
class Solution {
int countLeft(TreeNode *root) {
int cnt = 0;
for (; root; ++cnt, root = root->left);
return cnt;
}
int countRight(TreeNode *root) {
int cnt = 0;
for (; root; ++cnt, root = root->right);
return cnt;
}
public:
int countNodes(TreeNode* root) {
if (!root) return 0;
int left = countLeft(root), right = countRight(root);
if (left == right) return (1 << left) - 1;
return countNodes(root->left) + countNodes(root->right) + 1;
}
};

• # Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
def getHeight(self, root):
height = 0
while root:
height += 1
root = root.left
return height

def countNodes(self, root):
count = 0
while root:
l, r = map(self.getHeight, (root.left, root.right))
if l == r:
count += 2 ** l
root = root.right
else:
count += 2 ** r
root = root.left
return count