Maximum width of a binary tree
Given a binary tree, write a function to get the maximum width of the given tree. Width of a tree is maximum of widths of all levels.
Let us consider the below example tree.
1
/ \
2 3
/ \ \
4 5 8
/ \
6 7
For the above tree,
width of level 1 is 1,
width of level 2 is 2,
width of level 3 is 3
width of level 4 is 2.
So the maximum width of the tree is 3.
Method 1 (Using Level Order Traversal)
This method mainly involves two functions. One is to count nodes at a given level (getWidth), and other is to get the maximum width of the tree(getMaxWidth). getMaxWidth() makes use of getWidth() to get the width of all levels starting from root.
/*Function to print level order traversal of tree*/
getMaxWidth(tree)
maxWdth = 0
for i = 1 to height(tree)
width = getWidth(tree, i);
if(width > maxWdth)
maxWdth = width
return maxWidth
/*Function to get width of a given level */
getWidth(tree, level)
if tree is NULL then return 0;
if level is 1, then return 1;
else if level greater than 1, then
return getWidth(tree->left, level-1) +
getWidth(tree->right, level-1);
Below is the implementation of the above idea:
C++
// C++ program to calculate width of binary tree #include <bits/stdc++.h> using namespace std; /* A binary tree node has data, pointer to left child and a pointer to right child */class node { public: int data; node* left; node* right; }; /*Function protoypes*/int getWidth(node* root, int level); int height(node* node); node* newNode(int data); /* Function to get the maximum width of a binary tree*/int getMaxWidth(node* root) { int maxWidth = 0; int width; int h = height(root); int i; /* Get width of each level and compare the width with maximum width so far */ for (i = 1; i <= h; i++) { width = getWidth(root, i); if (width > maxWidth) maxWidth = width; } return maxWidth; } /* Get width of a given level */int getWidth(node* root, int level) { if (root == NULL) return 0; if (level == 1) return 1; else if (level > 1) return getWidth(root->left, level - 1) + getWidth(root->right, level - 1); } /* UTILITY FUNCTIONS *//* Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/int height(node* node) { if (node == NULL) return 0; else { /* compute the height of each subtree */ int lHeight = height(node->left); int rHeight = height(node->right); /* use the larger one */ return (lHeight > rHeight) ? (lHeight + 1) : (rHeight + 1); } } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */node* newNode(int data) { node* Node = new node(); Node->data = data; Node->left = NULL; Node->right = NULL; return (Node); } /* Driver code*/int main() { node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->right = newNode(8); root->right->right->left = newNode(6); root->right->right->right = newNode(7); /* Constructed bunary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 */ // Function call cout << "Maximum width is " << getMaxWidth(root) << endl; return 0; } // This code is contributed by rathbhupendra |
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C
// C program to calculate width of binary tree #include <stdio.h> #include <stdlib.h> /* A binary tree node has data, pointer to left child and a pointer to right child */struct node { int data; struct node* left; struct node* right; }; /*Function protoypes*/int getWidth(struct node* root, int level); int height(struct node* node); struct node* newNode(int data); /* Function to get the maximum width of a binary tree*/int getMaxWidth(struct node* root) { int maxWidth = 0; int width; int h = height(root); int i; /* Get width of each level and compare the width with maximum width so far */ for (i = 1; i <= h; i++) { width = getWidth(root, i); if (width > maxWidth) maxWidth = width; } return maxWidth; } /* Get width of a given level */int getWidth(struct node* root, int level) { if (root == NULL) return 0; if (level == 1) return 1; else if (level > 1) return getWidth(root->left, level - 1) + getWidth(root->right, level - 1); } /* UTILITY FUNCTIONS *//* Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/int height(struct node* node) { if (node == NULL) return 0; else { /* compute the height of each subtree */ int lHeight = height(node->left); int rHeight = height(node->right); /* use the larger one */ return (lHeight > rHeight) ? (lHeight + 1) : (rHeight + 1); } } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct node* newNode(int data) { struct node* node = (struct node*)malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return (node); } /* Driver code*/int main() { struct node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->right = newNode(8); root->right->right->left = newNode(6); root->right->right->right = newNode(7); /* Constructed bunary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 */ // Function call printf("Maximum width is %d \n", getMaxWidth(root)); getchar(); return 0; } |
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Java
// Java program to calculate width of binary tree /* A binary tree node has data, pointer to left child and a pointer to right child */class Node { int data; Node left, right; Node(int item) { data = item; left = right = null; } } class BinaryTree { Node root; /* Function to get the maximum width of a binary tree*/ int getMaxWidth(Node node) { int maxWidth = 0; int width; int h = height(node); int i; /* Get width of each level and compare the width with maximum width so far */ for (i = 1; i <= h; i++) { width = getWidth(node, i); if (width > maxWidth) maxWidth = width; } return maxWidth; } /* Get width of a given level */ int getWidth(Node node, int level) { if (node == null) return 0; if (level == 1) return 1; else if (level > 1) return getWidth(node.left, level - 1) + getWidth(node.right, level - 1); return 0; } /* UTILITY FUNCTIONS */ /* Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/ int height(Node node) { if (node == null) return 0; else { /* compute the height of each subtree */ int lHeight = height(node.left); int rHeight = height(node.right); /* use the larger one */ return (lHeight > rHeight) ? (lHeight + 1) : (rHeight + 1); } } /* Driver code */ public static void main(String args[]) { BinaryTree tree = new BinaryTree(); /* Constructed bunary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 */ tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.right = new Node(8); tree.root.right.right.left = new Node(6); tree.root.right.right.right = new Node(7); // Function call System.out.println("Maximum width is " + tree.getMaxWidth(tree.root)); } } // This code has been contributed by Mayank Jaiswal |
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Python
# Python program to find the maximum width of # binary tree using Level Order Traversal. # A binary tree node class Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None # Function to get the maximum width of a binary tree def getMaxWidth(root): maxWidth = 0 h = height(root) # Get width of each level and compare the # width with maximum width so far for i in range(1, h+1): width = getWidth(root, i) if (width > maxWidth): maxWidth = width return maxWidth # Get width of a given level def getWidth(root, level): if root is None: return 0 if level == 1: return 1 elif level > 1: return (getWidth(root.left, level-1) + getWidth(root.right, level-1)) # UTILITY FUNCTIONS # Compute the "height" of a tree -- the number of # nodes along the longest path from the root node # down to the farthest leaf node. def height(node): if node is None: return 0 else: # compute the height of each subtree lHeight = height(node.left) rHeight = height(node.right) # use the larger one return (lHeight+1) if (lHeight > rHeight) else (rHeight+1) # Driver code root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) root.right.right = Node(8) root.right.right.left = Node(6) root.right.right.right = Node(7) """ Constructed bunary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 """# Function call print "Maximum width is %d" % (getMaxWidth(root)) # This code is contributed by Naveen Aili |
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C#
// C# program to calculate width of binary tree using System; /* A binary tree node has data, pointer to left child and a pointer to right child */public class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; } } public class BinaryTree { public Node root; /* Function to get the maximum width of a binary tree*/ public virtual int getMaxWidth(Node node) { int maxWidth = 0; int width; int h = height(node); int i; /* Get width of each level and compare the width with maximum width so far */ for (i = 1; i <= h; i++) { width = getWidth(node, i); if (width > maxWidth) { maxWidth = width; } } return maxWidth; } /* Get width of a given level */ public virtual int getWidth(Node node, int level) { if (node == null) { return 0; } if (level == 1) { return 1; } else if (level > 1) { return getWidth(node.left, level - 1) + getWidth(node.right, level - 1); } return 0; } /* UTILITY FUNCTIONS */ /* Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/ public virtual int height(Node node) { if (node == null) { return 0; } else { /* compute the height of each subtree */ int lHeight = height(node.left); int rHeight = height(node.right); /* use the larger one */ return (lHeight > rHeight) ? (lHeight + 1) : (rHeight + 1); } } /* Driver code */ public static void Main(string[] args) { BinaryTree tree = new BinaryTree(); /* Constructed bunary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 */ tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.right = new Node(8); tree.root.right.right.left = new Node(6); tree.root.right.right.right = new Node(7); // Function call Console.WriteLine("Maximum width is " + tree.getMaxWidth(tree.root)); } } // This code is contributed by Shrikant13 |
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Output
Maximum width is 3
Time Complexity: O(n2) in the worst case.
We can use Queue based level order traversal to optimize the time complexity of this method. The Queue based level order traversal will take O(n) time in worst case. Thanks to Nitish, DivyaC and tech.login.id2 for suggesting this optimization. See their comments for implementation using queue based traversal.
Method 2 (Using Level Order Traversal with Queue)
In this method we store all the child nodes at the current level in the queue and then count the total number of nodes after the level order traversal for a particular level is completed. Since the queue now contains all the nodes of the next level, we can easily find out the total number of nodes in the next level by finding the size of queue. We then follow the same procedure for the successive levels. We store and update the maximum number of nodes found at each level.
Below is the implementation of the above idea:
C++
// A queue based C++ program to find maximum width // of a Binary Tree #include <bits/stdc++.h> using namespace std; /* A binary tree node has data, pointer to left child and a pointer to right child */struct Node { int data; struct Node* left; struct Node* right; }; // Function to find the maximum width of the tree // using level order traversal int maxWidth(struct Node* root) { // Base case if (root == NULL) return 0; // Initialize result int result = 0; // Do Level order traversal keeping track of number // of nodes at every level. queue<Node*> q; q.push(root); while (!q.empty()) { // Get the size of queue when the level order // traversal for one level finishes int count = q.size(); // Update the maximum node count value result = max(count, result); // Iterate for all the nodes in the queue currently while (count--) { // Dequeue an node from queue Node* temp = q.front(); q.pop(); // Enqueue left and right children of // dequeued node if (temp->left != NULL) q.push(temp->left); if (temp->right != NULL) q.push(temp->right); } } return result; } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct Node* newNode(int data) { struct Node* node = new Node; node->data = data; node->left = node->right = NULL; return (node); } // Driver code int main() { struct Node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->right = newNode(8); root->right->right->left = newNode(6); root->right->right->right = newNode(7); /* Constructed Binary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 */ // Function call cout << "Maximum width is " << maxWidth(root) << endl; return 0; } // This code is contributed by Nikhil Kumar // Singh(nickzuck_007) |
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Java
// Java program to calculate maximum width // of a binary tree using queue import java.util.LinkedList; import java.util.Queue; public class maxwidthusingqueue { /* A binary tree node has data, pointer to left child and a pointer to right child */ static class node { int data; node left, right; public node(int data) { this.data = data; } } // Function to find the maximum width of // the tree using level order traversal static int maxwidth(node root) { // Base case if (root == null) return 0; // Initialize result int maxwidth = 0; // Do Level order traversal keeping // track of number of nodes at every level Queue<node> q = new LinkedList<>(); q.add(root); while (!q.isEmpty()) { // Get the size of queue when the level order // traversal for one level finishes int count = q.size(); // Update the maximum node count value maxwidth = Math.max(maxwidth, count); // Iterate for all the nodes in // the queue currently while (count-- > 0) { // Dequeue an node from queue node temp = q.remove(); // Enqueue left and right children // of dequeued node if (temp.left != null) { q.add(temp.left); } if (temp.right != null) { q.add(temp.right); } } } return maxwidth; } // Function call public static void main(String[] args) { node root = new node(1); root.left = new node(2); root.right = new node(3); root.left.left = new node(4); root.left.right = new node(5); root.right.right = new node(8); root.right.right.left = new node(6); root.right.right.right = new node(7); /* Constructed Binary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 */ // Function call System.out.println("Maximum width = " + maxwidth(root)); } } // This code is contributed by Rishabh Mahrsee |
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Python
# Python program to find the maximum width of binary # tree using Level Order Traversal with queue. # A binary tree node class Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None # Function to get the maximum width of a binary tree def getMaxWidth(root): # base case if root is None: return 0 q = [] maxWidth = 0 q.insert(0, root) while (q != []): # Get the size of queue when the level order # traversal for one level finishes count = len(q) # Update the maximum node count value maxWidth = max(count, maxWidth) while (count is not 0): count = count-1 temp = q[-1] q.pop() if temp.left is not None: q.insert(0, temp.left) if temp.right is not None: q.insert(0, temp.right) return maxWidth # Driver program to test above function root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) root.right.right = Node(8) root.right.right.left = Node(6) root.right.right.right = Node(7) """ Constructed bunary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 """# Function call print "Maximum width is %d" % (getMaxWidth(root)) # This code is contributed by Naveen Aili |
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C#
// C# program to calculate maximum width // of a binary tree using queue using System; using System.Collections.Generic; public class maxwidthusingqueue { /* A binary tree node has data, pointer to left child and a pointer to right child */ public class node { public int data; public node left, right; public node(int data) { this.data = data; } } // Function to find the maximum width of // the tree using level order traversal static int maxwidth(node root) { // Base case if (root == null) return 0; // Initialize result int maxwidth = 0; // Do Level order traversal keeping // track of number of nodes at every level Queue<node> q = new Queue<node>(); q.Enqueue(root); while (q.Count != 0) { // Get the size of queue when the level order // traversal for one level finishes int count = q.Count; // Update the maximum node count value maxwidth = Math.Max(maxwidth, count); // Iterate for all the nodes in // the queue currently while (count-- > 0) { // Dequeue an node from queue node temp = q.Dequeue(); // Enqueue left and right children // of dequeued node if (temp.left != null) { q.Enqueue(temp.left); } if (temp.right != null) { q.Enqueue(temp.right); } } } return maxwidth; } // Driver code public static void Main(String[] args) { node root = new node(1); root.left = new node(2); root.right = new node(3); root.left.left = new node(4); root.left.right = new node(5); root.right.right = new node(8); root.right.right.left = new node(6); root.right.right.right = new node(7); /* Constructed Binary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 */ Console.WriteLine("Maximum width = " + maxwidth(root)); } } // This code is contributed by Princi Singh |
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Output
Maximum width is 3
Complexity Analysis:
Time Complexity: O(N) where N is the total number of nodes in the tree.In level order traversal, every node of the tree is processed once and hence the complexity due to the level order traversal is O(N) if there are total N nodes in the tree. Therefore, the time complexity is O(N).
Auxiliary Space : O(w) where w is the maximum width of the tree.
In level order traversal, a queue is maintained whose maximum size at any moment can go upto the maximum width of the binary tree.
Method 3 (Using Preorder Traversal)
In this method we create a temporary array count[] of size equal to the height of tree. We initialize all values in count as 0. We traverse the tree using preorder traversal and fill the entries in count so that the count array contains count of nodes at each level in Binary Tree.
Below is the implementation of the above idea:
C++
// C++ program to calculate width of binary tree #include <bits/stdc++.h> using namespace std; /* A binary tree node has data, pointer to left child and a pointer to right child */class node { public: int data; node* left; node* right; }; // A utility function to get // height of a binary tree int height(node* node); // A utility function to allocate // a new node with given data node* newNode(int data); // A utility function that returns // maximum value in arr[] of size n int getMax(int arr[], int n); // A function that fills count array // with count of nodes at every // level of given binary tree void getMaxWidthRecur(node* root, int count[], int level); /* Function to get the maximum width of a binary tree*/int getMaxWidth(node* root) { int width; int h = height(root); // Create an array that will // store count of nodes at each level int* count = new int[h]; int level = 0; // Fill the count array using preorder traversal getMaxWidthRecur(root, count, level); // Return the maximum value from count array return getMax(count, h); } // A function that fills count array // with count of nodes at every // level of given binary tree void getMaxWidthRecur(node* root, int count[], int level) { if (root) { count[level]++; getMaxWidthRecur(root->left, count, level + 1); getMaxWidthRecur(root->right, count, level + 1); } } /* UTILITY FUNCTIONS *//* Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/int height(node* node) { if (node == NULL) return 0; else { /* compute the height of each subtree */ int lHeight = height(node->left); int rHeight = height(node->right); /* use the larger one */ return (lHeight > rHeight) ? (lHeight + 1) : (rHeight + 1); } } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */node* newNode(int data) { node* Node = new node(); Node->data = data; Node->left = NULL; Node->right = NULL; return (Node); } // Return the maximum value from count array int getMax(int arr[], int n) { int max = arr[0]; int i; for (i = 0; i < n; i++) { if (arr[i] > max) max = arr[i]; } return max; } /* Driver code*/int main() { node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->right = newNode(8); root->right->right->left = newNode(6); root->right->right->right = newNode(7); cout << "Maximum width is " << getMaxWidth(root) << endl; return 0; } // This is code is contributed by rathbhupendra |
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C
// C program to calculate width of binary tree #include <stdio.h> #include <stdlib.h> /* A binary tree node has data, pointer to left child and a pointer to right child */struct node { int data; struct node* left; struct node* right; }; // A utility function to get height of a binary tree int height(struct node* node); // A utility function to allocate a new node with given data struct node* newNode(int data); // A utility function that returns maximum value in arr[] of // size n int getMax(int arr[], int n); // A function that fills count array with count of nodes at // every level of given binary tree void getMaxWidthRecur(struct node* root, int count[], int level); /* Function to get the maximum width of a binary tree*/int getMaxWidth(struct node* root) { int width; int h = height(root); // Create an array that will store count of nodes at // each level int* count = (int*)calloc(sizeof(int), h); int level = 0; // Fill the count array using preorder traversal getMaxWidthRecur(root, count, level); // Return the maximum value from count array return getMax(count, h); } // A function that fills count array with count of nodes at // every level of given binary tree void getMaxWidthRecur(struct node* root, int count[], int level) { if (root) { count[level]++; getMaxWidthRecur(root->left, count, level + 1); getMaxWidthRecur(root->right, count, level + 1); } } /* UTILITY FUNCTIONS *//* Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/int height(struct node* node) { if (node == NULL) return 0; else { /* compute the height of each subtree */ int lHeight = height(node->left); int rHeight = height(node->right); /* use the larger one */ return (lHeight > rHeight) ? (lHeight + 1) : (rHeight + 1); } } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct node* newNode(int data) { struct node* node = (struct node*)malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return (node); } // Return the maximum value from count array int getMax(int arr[], int n) { int max = arr[0]; int i; for (i = 0; i < n; i++) { if (arr[i] > max) max = arr[i]; } return max; } /* Driver program to test above functions*/int main() { struct node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->right = newNode(8); root->right->right->left = newNode(6); root->right->right->right = newNode(7); /* Constructed bunary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 */ printf("Maximum width is %d \n", getMaxWidth(root)); getchar(); return 0; } |
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Java
// Java program to calculate width of binary tree /* A binary tree node has data, pointer to left child and a pointer to right child */class Node { int data; Node left, right; Node(int item) { data = item; left = right = null; } } class BinaryTree { Node root; /* Function to get the maximum width of a binary tree*/ int getMaxWidth(Node node) { int width; int h = height(node); // Create an array that will store count of nodes at // each level int count[] = new int[10]; int level = 0; // Fill the count array using preorder traversal getMaxWidthRecur(node, count, level); // Return the maximum value from count array return getMax(count, h); } // A function that fills count array with count of nodes // at every level of given binary tree void getMaxWidthRecur(Node node, int count[], int level) { if (node != null) { count[level]++; getMaxWidthRecur(node.left, count, level + 1); getMaxWidthRecur(node.right, count, level + 1); } } /* UTILITY FUNCTIONS */ /* Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/ int height(Node node) { if (node == null) return 0; else { /* compute the height of each subtree */ int lHeight = height(node.left); int rHeight = height(node.right); /* use the larger one */ return (lHeight > rHeight) ? (lHeight + 1) : (rHeight + 1); } } // Return the maximum value from count array int getMax(int arr[], int n) { int max = arr[0]; int i; for (i = 0; i < n; i++) { if (arr[i] > max) max = arr[i]; } return max; } /* Driver program to test above functions */ public static void main(String args[]) { BinaryTree tree = new BinaryTree(); /* Constructed bunary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 */ tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.right = new Node(8); tree.root.right.right.left = new Node(6); tree.root.right.right.right = new Node(7); System.out.println("Maximum width is " + tree.getMaxWidth(tree.root)); } } // This code has been contributed by Mayank Jaiswal |
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Python
# Python program to find the maximum width of # binary tree using Preorder Traversal. # A binary tree node class Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None # Function to get the maximum width of a binary tree def getMaxWidth(root): h = height(root) # Create an array that will store count of nodes at each level count = [0] * h level = 0 # Fill the count array using preorder traversal getMaxWidthRecur(root, count, level) # Return the maximum value from count array return getMax(count, h) # A function that fills count array with count of nodes at every # level of given binary tree def getMaxWidthRecur(root, count, level): if root is not None: count[level] += 1 getMaxWidthRecur(root.left, count, level+1) getMaxWidthRecur(root.right, count, level+1) # UTILITY FUNCTIONS # Compute the "height" of a tree -- the number of # nodes along the longest path from the root node # down to the farthest leaf node. def height(node): if node is None: return 0 else: # compute the height of each subtree lHeight = height(node.left) rHeight = height(node.right) # use the larger one return (lHeight+1) if (lHeight > rHeight) else (rHeight+1) # Return the maximum value from count array def getMax(count, n): max = count[0] for i in range(1, n): if (count[i] > max): max = count[i] return max # Driver program to test above function root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) root.right.right = Node(8) root.right.right.left = Node(6) root.right.right.right = Node(7) """ Constructed bunary tree is: 1 / \ 2 3 / \ \ 4 5 8 / \ 6 7 """ print "Maximum width is %d" % (getMaxWidth(root)) # This code is contributed by Naveen Aili |
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C#
// C# program to calculate width of binary tree using System; /* A binary tree node has data, pointer to left child and a pointer to right child */public class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; } } public class BinaryTree { Node root; /* Function to get the maximum width of a binary tree*/ int getMaxWidth(Node node) { int width; int h = height(node); // Create an array that will store // count of nodes at each level int[] count = new int[10]; int level = 0; // Fill the count array using preorder traversal getMaxWidthRecur(node, count, level); // Return the maximum value from count array return getMax(count, h); } // A function that fills count // array with count of nodes at every // level of given binary tree void getMaxWidthRecur(Node node, int[] count, int level) { if (node != null) { count[level]++; getMaxWidthRecur(node.left, count, level + 1); getMaxWidthRecur(node.right, count, level + 1); } } /* UTILITY FUNCTIONS */ /* Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/ int height(Node node) { if (node == null) return 0; else { /* compute the height of each subtree */ int lHeight = height(node.left); int rHeight = height(node.right); /* use the larger one */ return (lHeight > rHeight) ? (lHeight + 1) : (rHeight + 1); } } // Return the maximum value from count array int getMax(int[] arr, int n) { int max = arr[0]; int i; for (i = 0; i < n; i++) { if (arr[i] > max) max = arr[i]; } return max; } /* Driver program to test above functions */ public static void Main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.right = new Node(8); tree.root.right.right.left = new Node(6); tree.root.right.right.right = new Node(7); Console.WriteLine("Maximum width is " + tree.getMaxWidth(tree.root)); } } // This code is contributed Rajput-Ji |
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Output
Maximum width is 3
Time Complexity: O(n)
Thanks to Raja and jagdish for suggesting this method.
Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.
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