Iterative diagonal traversal of binary tree
Consider lines of slope -1 passing between nodes. Given a Binary Tree, print all diagonal elements in a binary tree belonging to same line.
Input : Root of below treeOutput : Diagonal Traversal of binary tree : 8 10 14 3 6 7 13 1 4
We have discussed recursive solution in below post.
Diagonal Traversal of Binary Tree
In this post, iterative solution is discussed. The idea is to use a queue to store only the left child of current node. After printing the data of current node make the current node to its right child, if present.
A delimiter NULL is used to mark the starting of next diagonal.
Below is the implementation of above approach.
C++
/* C++ program to construct string from 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; Node *left, *right; }; /* Helper function that allocates a new node */Node* newNode(int data) { Node* node = (Node*)malloc(sizeof(Node)); node->data = data; node->left = node->right = NULL; return (node); } // Iterative function to print diagonal view void diagonalPrint(Node* root) { // base case if (root == NULL) return; // inbuilt queue of Treenode queue<Node*> q; // push root q.push(root); // push delimiter q.push(NULL); while (!q.empty()) { Node* temp = q.front(); q.pop(); // if current is delimiter then insert another // for next diagonal and cout nextline if (temp == NULL) { // if queue is empty return if (q.empty()) return; // output nextline cout << endl; // push delimiter again q.push(NULL); } else { while (temp) { cout << temp->data << " "; // if left child is present // push into queue if (temp->left) q.push(temp->left); // current equals to right child temp = temp->right; } } } } // Driver Code int main() { Node* root = newNode(8); root->left = newNode(3); root->right = newNode(10); root->left->left = newNode(1); root->left->right = newNode(6); root->right->right = newNode(14); root->right->right->left = newNode(13); root->left->right->left = newNode(4); root->left->right->right = newNode(7); diagonalPrint(root); } |
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Java
// Java program to con string from binary tree import java.util.*; class solution { // A binary tree node has data, pointer to left // child and a pointer to right child static class Node { int data; Node left, right; }; // Helper function that allocates a new node static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = node.right = null; return (node); } // Iterative function to print diagonal view static void diagonalPrint(Node root) { // base case if (root == null) return; // inbuilt queue of Treenode Queue<Node> q= new LinkedList<Node>(); // add root q.add(root); // add delimiter q.add(null); while (q.size()>0) { Node temp = q.peek(); q.remove(); // if current is delimiter then insert another // for next diagonal and cout nextline if (temp == null) { // if queue is empty return if (q.size()==0) return; // output nextline System.out.println(); // add delimiter again q.add(null); } else { while (temp!=null) { System.out.print( temp.data + " "); // if left child is present // add into queue if (temp.left!=null) q.add(temp.left); // current equals to right child temp = temp.right; } } } } // Driver Code public static void main(String args[]) { Node root = newNode(8); root.left = newNode(3); root.right = newNode(10); root.left.left = newNode(1); root.left.right = newNode(6); root.right.right = newNode(14); root.right.right.left = newNode(13); root.left.right.left = newNode(4); root.left.right.right = newNode(7); diagonalPrint(root); } } //contributed by Arnab Kundu |
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Python3
# Python3 program to construct string from binary tree class Node: def __init__(self,data): self.val = data self.left = None self.right = None # Function to print diagonal view def diagonalprint(root): # base case if root is None: return # queue of treenode q = [] # Append root q.append(root) # Append delimiter q.append(None) while len(q) > 0: temp = q.pop(0) # If current is delimiter then insert another # for next diagonal and cout nextline if not temp: # If queue is empty then return if len(q) == 0: return # Print output on nextline print(' ') # append delimiter again q.append(None) else: while temp: print(temp.val, end = ' ') # If left child is present # append into queue if temp.left: q.append(temp.left) # current equals to right child temp = temp.right # Driver Code root = Node(8) root.left = Node(3) root.right = Node(10) root.left.left = Node(1) root.left.right = Node(6) root.right.right = Node(14) root.right.right.left = Node(13) root.left.right.left = Node(4) root.left.right.right = Node(7) diagonalprint(root) # This code is contributed by Praveen kumar |
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C#
// C# program to con string from binary tree using System; using System.Collections; class GFG { // 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; }; // Helper function that // allocates a new node static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = node.right = null; return (node); } // Iterative function to print diagonal view static void diagonalPrint(Node root) { // base case if (root == null) return; // inbuilt queue of Treenode Queue q = new Queue(); // Enqueue root q.Enqueue(root); // Enqueue delimiter q.Enqueue(null); while (q.Count > 0) { Node temp = (Node) q.Peek(); q.Dequeue(); // if current is delimiter then insert another // for next diagonal and cout nextline if (temp == null) { // if queue is empty return if (q.Count == 0) return; // output nextline Console.WriteLine(); // Enqueue delimiter again q.Enqueue(null); } else { while (temp != null) { Console.Write( temp.data + " "); // if left child is present // Enqueue into queue if (temp.left != null) q.Enqueue(temp.left); // current equals to right child temp = temp.right; } } } } // Driver Code public static void Main(String []args) { Node root = newNode(8); root.left = newNode(3); root.right = newNode(10); root.left.left = newNode(1); root.left.right = newNode(6); root.right.right = newNode(14); root.right.right.left = newNode(13); root.left.right.left = newNode(4); root.left.right.right = newNode(7); diagonalPrint(root); } } // This code is contributed by Arnab Kundu |
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Output:
8 10 14 3 6 7 13 1 4
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