You have been given a Binary Tree of integers. You are supposed to return the maximum width of the given Binary Tree. The maximum width of the tree is the maximum width among all the levels of the given tree.
The width of one level is defined as the length between the leftmost and the rightmost, non-null nodes in the level, where the null nodes in between the leftmost and rightmost are excluded into length calculation.
For the given binary tree
The maximum width will be at the third level with the length of 3, i.e. {4, 5, 6}.
The only line of input contains elements in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place.
For example, the input for the tree depicted in the below image would be :
1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1
Explanation :
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)
The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null(-1).
The above format was just to provide clarity on how the input is formed for a given tree.
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
Print a single line that contains a single integer that denotes the maximum width for the given tree.
You do not need to print anything; it has already been taken care of. Just implement the given function.
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
3
The maximum width will be at the third level with the length of 3, i.e. {4, 5, 6}.
2 7 5 2 6 -1 9 -1 -1 5 11 4 -1 -1 -1 -1 -1 -1 -1
3
0 <= 'N' <= 5 * 10 ^ 5
0 <= 'DATA' <= 10 ^ 6 and data != -1
Where ‘N’ is the total number of nodes in the binary tree, and 'DATA' is the value of the binary tree node.
Time Limit: 1sec.
How can you find the number of nodes in each level of the given tree?
The straightforward intuition is that first, find the maximum levels possible in the given tree, called the "HEIGHT" of the tree. And then, for each level in height, we will find the number of nodes in each level. And the maximum number of the node among all levels will be the maximum width of the given binary tree. So the implementation of our intuition takes bellow steps:
2. We will go for each level in [1: "HEIGHT"] and find how many nodes it contains at that level. The maximum number of nodes among all levels will be the maximum width of the given tree. For finding the number of nodes at any level, let's say "LEVEL", Steps are as follows:
3. The maximum number of nodes among all levels will be the maximum width of the given tree.
O(N * H), where ‘N’ is the number of nodes in the given binary tree, and ‘H’ is the height of the binary tree.
We are finding the height of the binary tree, which will take cost O(H), in the worst case (Skewed Trees), ‘H’ is equal to ‘N’. And for each level in height, we will find the number of nodes in each level. So there can be at most ‘H’ level, and for each level, we will find the number of nodes that can take O(N) cost. So overall time complexity will be O(N * H).
O(H), where ‘H’ is the height of the binary tree.
We are using a recursive algorithm to find the height of the tree and find the number of nodes in each level. For getting the number of nodes for each level, the level will be at most the height of the tree. So in the worst case (Skewed Trees), 'H' is equal to 'N', there can be at most 'N' height for which we will have to push function 'N' time into the stack. So overall space complexity would be O(H).
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Interview problems
C++ solution
int getMaxWidth(TreeNode<int> *root)
{
if(root==NULL) return 0;
queue<TreeNode<int>*>q;
int res=INT_MIN;
q.push(root);
while(q.empty()==false)
{
int ct=q.size();
res=max(res,ct);
for(int i=0;i<ct;i++)
{
TreeNode<int>*curr=q.front();
q.pop();
if(curr->left !=NULL)
q.push(curr->left);
if(curr->right !=NULL)
q.push(curr->right);
}
}
return res;
}
Interview problems
C++ || Level order
int getMaxWidth(TreeNode<int> *root){
if(!root) return 0;
int ans = 0;
queue<TreeNode<int>*> q;
q.push(root);
while(!q.empty()){
int n = q.size();
ans = max(ans, n);
for(int i=0; i<n; i++){
TreeNode<int>* front = q.front();
q.pop();
if(front->left)
q.push(front->left);
if(front->right)
q.push(front->right);
}
}
return ans;
}
Interview problems
Best Solution using BFS traversal
/*
Time Complexity : O(N)
Space Complexity : O(N)
*/
/************************************************************
Following is the TreeNode class structure
template <typename T>
class TreeNode {
public:
T val;
TreeNode<T> *left;
TreeNode<T> *right;
TreeNode(T val) {
this->val = val;
left = NULL;
right = NULL;
}
};
************************************************************/
int getMaxWidth(TreeNode<int> *root)
{
if(!root)return 0;
queue<TreeNode<int>*> q;
q.push(root);
int maxi=0;
while(!q.empty()){
int size=q.size();
vector<int> row;
while(size--){
TreeNode<int>* node=q.front();
q.pop();
row.push_back(node->val);
if(node->left)q.push(node->left);
if(node->right)q.push(node->right);
}
int n=row.size();
maxi=max(maxi,n);
}
return maxi;
}
Interview problems
Python working code
class Pair:
def __init__(self,node,num):
self.node=node
self.num=num
def getMaxWidth(root):
if root is None:
return 0
ans = 0
q = deque()
q.append(Pair(root, 0))
while q:
size = len(q)
mmin = q[0].num
first, last = 0, 0
for i in range(size):
cur_pair = q.popleft()
cur_id = cur_pair.num - mmin
node = cur_pair.node
if i == 0:
first = cur_id
if i == size - 1:
last = cur_id
if node.left:
q.append(Pair(node.left, cur_id * 2 + 1))
if node.right:
q.append(Pair(node.right, cur_id * 2 + 2))
ans = max(ans, last - first + 1)
return ans
Interview problems
Simple C++ solution using lambda function
int getMaxWidth(TreeNode<int> *root)
{
if(!root){
return 0;
}
function <int(TreeNode<int> *)>getWidth=[&](TreeNode<int> *root){
queue<TreeNode<int> *>q;
int ans = 0;
q.push(root);
while(!q.empty()){
auto k = q.front();
int size = q.size();
ans = max(ans,size);
for(int i=0;i<size;i++){
auto it = q.front();
if(it->left){
q.push(it->left);
}
if(it->right){
q.push(it->right);
}
q.pop();
}
}
return ans;
};
return getWidth(root);
}Interview problems
Simple c++ soln, TC-O(N), S.C-O(W)
int getMaxWidth(TreeNode<int> *root){
if(!root) return 0;
queue<TreeNode<int>*> q;
q.push(root);
q.push(NULL);
int ans=0,cnt=0;
while(!q.empty()){
TreeNode<int>* temp=q.front();
q.pop();
if(temp==NULL){
ans=max(ans,cnt);
cnt=0;
if(!q.empty()) q.push(NULL);
}
else {
cnt++;
if (temp->left) q.push(temp->left);
if (temp->right) q.push(temp->right);
}
}
return ans;
}
Interview problems
simple java solution using Level order traversal
import java.util.*;
// import javax.swing.tree.TreeNode;
public class Solution {
public static int getMaxWidth(TreeNode root) {
if (root == null) return 0;
Queue<TreeNode> queue= new LinkedList();
queue.offer(root);
// TreeSet<Integer> l = new TreeSet();
int max = Integer.MIN_VALUE;
while(!queue.isEmpty()) {
int size = queue.size();
List<Integer> ll = new ArrayList<>();
for (int i=0;i<size;i++) {
if (queue.peek().left !=null) {
queue.offer(queue.peek().left);
}
if (queue.peek().right != null) {
queue.offer(queue.peek().right);
}
ll.add(queue.poll().val);
}
max = Math.max(max,ll.size());
}
return max;
}
}
Interview problems
C++ Easy Solution Using Queue
int getMaxWidth(TreeNode<int> *root)
{
// Write your code here.
if(root == NULL) return 0;
int count = 0, ans = INT_MIN;
queue<TreeNode<int> *> q;
q.push(root);
q.push(NULL);
while(!q.empty()){
TreeNode<int> *top = q.front();
q.pop();
if(top == NULL){
if(q.empty()) return ans = max(ans,count);
ans = max(ans,count);
count = 0;
q.push(NULL);
}
else{
count++;
if(top->left) q.push(top->left);
if(top->right) q.push(top->right);
}
}
return ans;
}Interview problems
c++ #levelordertraversal
int getMaxWidth(TreeNode<int> *root)
{
// Write your code here.
queue<TreeNode<int>*>q;
q.push(root);
int maxi =INT_MIN;
if(root==NULL)
{
return 0;
}
while(!q.empty())
{
int size =q.size();
int count =0;
for(int i=0;i<size;i++)
{
TreeNode<int>*top=q.front();q.pop();
count++;
if(top->left)
{
q.push(top->left);
}
if(top->right)
{
q.push(top->right);
}
}
maxi=max(maxi,count);
}
return maxi;
}
Interview problems
python
from queue import Queue
def getMaxWidth(root):
# Write your code here.
if not root:
return 0
max_width = float('-inf')
q = Queue()
q.put(root)
while not q.empty():
level_width = q.qsize()
max_width = max(max_width, level_width)
for i in range(level_width):
current = q.get()
if current.left:
q.put(current.left)
if current.right:
q.put(current.right)
return max_width
pass
