Created
July 2, 2017 18:24
-
-
Save anil477/4c55a3fa18373fabcd4f19f74eaef024 to your computer and use it in GitHub Desktop.
Boundary Traversal of binary tree
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
// http://www.geeksforgeeks.org/boundary-traversal-of-binary-tree/ | |
class Node | |
{ | |
int data; | |
Node left, right; | |
Node(int item) | |
{ | |
data = item; | |
left = right = null; | |
} | |
} | |
class BinaryTree | |
{ | |
Node root; | |
// A simple function to print leaf nodes of a binary tree | |
// This method is same as inorder traversal except that we need only the leaf nodes. | |
void printLeaves(Node node) | |
{ | |
if (node != null) | |
{ | |
printLeaves(node.left); | |
// Print it if it is a leaf node | |
if (node.left == null && node.right == null) | |
System.out.print(node.data + " "); | |
printLeaves(node.right); | |
} | |
} | |
// A function to print all left boundry nodes, except a leaf node. | |
// Print the nodes in TOP DOWN manner - note the data is printed before the recursive call | |
// Initially this method is called with the root.left and root | |
// NOTE - the boundary methods don't print the leaf node. | |
void printBoundaryLeft(Node node) | |
{ | |
if (node != null) | |
{ | |
if (node.left != null) | |
{ | |
// to ensure top down order, print the node before calling itself for left subtree | |
System.out.print(node.data + " "); | |
printBoundaryLeft(node.left); | |
} | |
else if (node.right != null) | |
{ | |
// if the tree does not have left nodes, then right nodes will form the boundary. | |
// This is taken here by this if condition. | |
// if there's no left-subtree but only right sub-tree and right-subtree has both left and right subtree, | |
// then left subtree should be printed. | |
// that's why the left condition is above the right condition | |
System.out.print(node.data + " "); | |
printBoundaryLeft(node.right); | |
} | |
// do nothing if it is a leaf node, this way we avoid | |
// duplicates in output | |
} | |
} | |
// A function to print all right boundry nodes, except a leaf node | |
// Print the nodes in BOTTOM UP manner - note the data is printed after the recursive call | |
// Initially this method is called with the root.right and root | |
// NOTE - the boundary methods don't print the leaf node. | |
void printBoundaryRight(Node node) | |
{ | |
if (node != null) | |
{ | |
if (node.right != null) | |
{ | |
// to ensure bottom up order, first call for right subtree, then print this node | |
printBoundaryRight(node.right); | |
System.out.print(node.data + " "); | |
} | |
else if (node.left != null) | |
{ | |
// if the tree does not have right nodes, then left nodes will form the boundary. | |
// This is taken here by this if condition. | |
// if there's no right-subtree but only left sub-tree and left-subtree has both left and right subtree, | |
// then right subtree should be printed. | |
// that's why the right condition is above the left condition | |
printBoundaryRight(node.left); | |
System.out.print(node.data + " "); | |
} | |
// do nothing if it is a leaf node, this way we avoid | |
// duplicates in output | |
} | |
} | |
// A function to do boundary traversal of a given binary tree | |
void printBoundary(Node node) | |
{ | |
if (node != null) | |
{ | |
// the root data is printed explicity and not included in any method as it will overlap | |
System.out.print(node.data + " "); | |
// Since it's required to print data in anti-clock wise direction, the following order is chosen | |
// Print the left boundary in top-down manner. | |
printBoundaryLeft(node.left); | |
// Print all leaf nodes | |
printLeaves(node.left); | |
printLeaves(node.right); | |
// Print the right boundary in bottom-up manner | |
printBoundaryRight(node.right); | |
} | |
} | |
// Driver program to test above functions | |
public static void main(String args[]) | |
{ | |
BinaryTree tree = new BinaryTree(); | |
tree.root = new Node(20); | |
tree.root.left = new Node(8); | |
tree.root.left.left = new Node(4); | |
tree.root.left.right = new Node(12); | |
tree.root.left.right.left = new Node(10); | |
tree.root.left.right.right = new Node(14); | |
tree.root.right = new Node(22); | |
tree.root.right.right = new Node(25); | |
tree.printBoundary(tree.root); | |
} | |
} |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment