Leetcode Count Complete Tree Nodes Java Solution

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 2h nodes inclusive at the last level h.

Example:

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

Output: 6```

Solution 1:

``````class HackerHeap {
public int countNodes(TreeNode root) {
if(root == null) return 0;
int cnodes = 0;
while(!q.isEmpty()){
TreeNode curr = q.poll();
if(curr.left!=null){
cnodes++;
}else if(curr.right==null){
cnodes++;
}else{
}
}
return cnodes;
}
}

``````

Solution 2:

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class HackerHeap {

public int countNodes(TreeNode root) {
// if the tree is empty
if (root == null) return 0;

int d = computeDepth(root);
// if the tree contains 1 node
if (d == 0) return 1;

// Last level nodes are enumerated from 0 to 2**d - 1 (left -> right).
// Perform binary search to check how many nodes exist.
int left = 1, right = (int)Math.pow(2, d) - 1;
int pivot;
while (left <= right) {
pivot = left + (right - left) / 2;
if (exists(pivot, d, root)) left = pivot + 1;
else right = pivot - 1;
}

// The tree contains 2**d - 1 nodes on the first (d - 1) levels
// and left nodes on the last level.
return (int)Math.pow(2, d) - 1 + left;
}
public int computeDepth(TreeNode node) {
int d = 0;
while (node.left != null) {
node = node.left;
++d;
}
return d;
}

// Last level nodes are enumerated from 0 to 2**d - 1 (left -> right).
// Return True if last level node idx exists.
// Binary search with O(d) complexity.
public boolean exists(int idx, int d, TreeNode node) {
int left = 0, right = (int)Math.pow(2, d) - 1;
int pivot;
for(int i = 0; i < d; ++i) {
pivot = left + (right - left) / 2;
if (idx <= pivot) {
node = node.left;
right = pivot;
}
else {
node = node.right;
left = pivot + 1;
}
}
return node != null;
}
}``````