Algorithm
Problem Name: 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.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
void dfs(struct TreeNode *node, int i, int n, int *x) {
if (i == n) {
if (node->left) (*x) ++;
if (node->right) (*x) ++;
return;
}
if ((*x) % 2) return;
dfs(node->left, i + 1, n, x);
if ((*x) % 2) return;
dfs(node->right, i + 1, n, x);
}
int countNodes(struct TreeNode* root) {
struct TreeNode *node = root;
int i = 0;
int x = 0;
if (!node) return 0;
while (node) {
i ++;
node = node->right;
}
dfs(root, 1, i, &x);
return (1 << i) - 1 + x;
}
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int countNodes(TreeNode root) {
int leftDepth = getDepth(root, -1);
int rightDepth = getDepth(root, 1);
if (leftDepth == rightDepth) {
return (1 << leftDepth) - 1;
}
return 1 + countNodes(root.left) + countNodes(root.right);
}
private int getDepth(TreeNode root, int dir) {
int depth = 0;
while (root != null) {
depth++;
if (dir == -1) {
root = root.left;
} else {
root = root.right;
}
}
return depth;
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const countNodes = function(root) {
if (root == null) return 0;
const payload = { depth: 0, numOfLast: 0, total: 0 };
traverse([root], 0, payload);
return payload.total;
};
function traverse(row, depth, obj) {
const next = [];
for (let i = 0; i < row.length; i++) {
if (row[i].left) next.push(row[i].left);
if (row[i].right) next.push(row[i].right);
}
if (Math.pow(2, depth + 1) !== next.length) {
obj.total = Math.pow(2, depth + 1) - 1 + next.length;
return;
}
if (next.length) traverse(next, depth + 1, obj);
}
Input
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def countNodes(self, root):
if not root: return 0
l = self.getDepth(root.left)
r = self.getDepth(root.right)
if l == r:
return (1 << l) + self.countNodes(root.right)
return (1 << r) + self.countNodes(root.left)
def getDepth(self, root):
if not root: return 0
return 1 + self.getDepth(root.left)
Input
#5 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _0222_CountCompleteTreeNodes
{
public int CountNodes(TreeNode root)
{
if (root == null) { return 0; }
// Find Depth
var current = root;
var depth = 0;
while (current.left != null)
{
current = current.left;
depth++;
}
if (depth == 0) return 1;
var count = (int)Math.Pow(2, depth) - 1;
var l = 0;
var r = count;
while (l < = r)
{
var mid = l + (r - l) / 2;
if (FindNode(root, mid, count, depth))
l = mid + 1;
else
r = mid - 1;
}
return count + l;
}
public bool FindNode(TreeNode root, int index, int count, int depth)
{
var l = 0;
var r = count;
for (int i = 0; i < depth; i++)
{
var mid = l + (r - l) / 2;
if (index < = mid)
{
root = root.left;
r = mid;
}
else
{
root = root.right;
l = mid + 1;
}
}
return root != null;
}
}
}
Input
Output