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# Count Complete Tree Nodes LeetCode Solution

## Problem – Count Complete Tree Nodes LeetCode Solution

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 = 
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.

## Count Complete Tree Nodes LeetCode Solution in Java

``````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);
}
}
``````

## Count Complete Tree Nodes LeetCode Solution in C++

``````class Solution {

public:

int countNodes(TreeNode* root) {

if(!root) return 0;

int hl=0, hr=0;

TreeNode *l=root, *r=root;

while(l) {hl++;l=l->left;}

while(r) {hr++;r=r->right;}

if(hl==hr) return pow(2,hl)-1;

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

}

};
``````

## Count Complete Tree Nodes LeetCode Solution in Python

`````` class Solution:
# @param {TreeNode} root
# @return {integer}
def countNodes(self, root):
if not root:
return 0
leftDepth = self.getDepth(root.left)
rightDepth = self.getDepth(root.right)
if leftDepth == rightDepth:
return pow(2, leftDepth) + self.countNodes(root.right)
else:
return pow(2, rightDepth) + self.countNodes(root.left)

def getDepth(self, root):
if not root:
return 0
return 1 + self.getDepth(root.left)
``````
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