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Tree traversal

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ap_Cooperative_dev answered on May 27, 2022 Popularity 10/10 Helpfulness 2/10

answer Tree traversal

related preorder traversal

related How to perform in-order traversal of a binary tree?

related traversal tree

related Tree: Postorder Traversal

Tree traversal

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A tree is a graph whose degree of node== # of it's children & is acyclic 
Binary tree is a tree with each node having atmost 2 children.
2 ways to traverse each node once:
BFS - level wise
DFS - BY RECURSIVELY TRAVERSING ROOT,LEFT SUBTREE (L) & RIGHT SUBTREE (R)
NOTE: THERE ARE 3! WAYS OF DOING A DFS, BASED ON ORDER OF Root,L,R.     
but only 3 are useful :
Root,L,R- preorder traversal 
L,Root,R- inorder traversal 
L,R,Root- postorder traversal

Popularity 10/10 Helpfulness 2/10 Language whatever

Source: Grepper

Tags: inorder post tree tree-traversal whatever

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Contributed on Aug 03 2022

ap_Cooperative_dev

0 Answers Avg Quality 2/10

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preorder traversal

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<> Preorder		Root -> Left -> Right 
<> Inorder		Left -> Root -> Right
<> Postorder	Left -> Right -> Root

Popularity 10/10 Helpfulness 4/10 Language javascript

Source: Grepper

Tags: depth-first-search depth-first-searc

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Contributed on Jan 09 2023

correctYourPosture

0 Answers Avg Quality 2/10

How to perform in-order traversal of a binary tree?

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/*
	This is an implementation that collects the 
	values of the nodes of a binary tree by performing 
	an in-order traversal of the tree.

	Let n be the number of binary tree nodes
	Time complexity: O(n) 
	Space complexity: O(n)
*/
import java.util.List;
import java.util.ArrayList;

public class BTInOrderTraversal {
	private BTNode BTRoot;
	public BTInOrderTraversal() {
		/*
		 * Create tree below:
		 * 1
		 * \
		 * 2
		 * /
		 * 3
		 */
		BTRoot = new BTNode(1, null, null);
		BTNode rootRight = new BTNode(2, null, null);
		BTRoot.right = rootRight;
		BTNode rootRightLeft = new BTNode(3, null, null);
		rootRight.left = rootRightLeft;
	}
	public static void main(String[] args) {
		BTInOrderTraversal application = new BTInOrderTraversal();
		List<Integer> values = application.inorderTraversal();
		System.out.println(values); // [1, 3, 2]
	}
	// Perform in-order traversal through the tree.
	public List<Integer> inorderTraversal() {
		List<Integer> list = new ArrayList<>();
		populateList(BTRoot, list);
		return list;
	}
	// Helper method to populate list by performing
	// in-order traversal through the tree.
	private void populateList(BTNode root, List<Integer> list) {
		if (root == null) {
			return;
		}
		if (root.left != null) {
			populateList(root.left, list);
		}
		list.add(root.val);
		if (root.right != null) {
			populateList(root.right, list);
		}
	}
	// Class representing a binary tree node
	// with pointers to value, left, and right nodes
	private class BTNode {
		int val;
		BTNode left;
		BTNode right;
		public BTNode(int val, BTNode left, BTNode right) {
			this.val = val;
			this.left = left;
			this.right = right;
		}
	}
}

Popularity 10/10 Helpfulness 2/10 Language java

Source: Grepper

Tags: java

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Contributed on May 27 2022

Wissam

0 Answers Avg Quality 2/10

traversal tree

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class Node:
     def __init__(self,data):
          self.data = data
          self.parent = None
          self.left = None
          self.right = None

     def __repr__(self):
          return repr(self.data)

     def add_left(self,node):
         self.left = node
         if node is not None:
             node.parent = self

     def add_right(self,node):
         self.right = node
         if node is not None:
             node.parent = self
'''
Example:
          _2_
        /       \
       7         9
      / \         \
     1   6         8
         / \       / \
       5   10   3   4
'''
def create_tree():
    two = Node(2)
    seven = Node(7)
    nine = Node(9)
    two.add_left(seven)
    two.add_right(nine)
    one = Node(1)
    six = Node(6)
    seven.add_left(one)
    seven.add_right(six)
    five = Node(5)
    ten = Node(10)
    six.add_left(five)
    six.add_right(ten)
    eight = Node(8)
    three = Node(3)
    four = Node(4)
    nine.add_right(eight)
    eight.add_left(three)
    eight.add_right(four)

    # now return the root node
    return two

def pre_order(node):
    print(node)
    if node.left:
        pre_order(node.left)
    if node.right:
        pre_order(node.right)

def in_order(node):
    if node.left:
        in_order(node.left)
    print(node)
    if node.right:
        in_order(node.right)

def post_order(node):
    if node.left:
        post_order(node.left)
    if node.right:
        post_order(node.right)
    print(node)

if __name__ == "__main__":
    root = create_tree()
    print("\nPre-order traversal:")
    pre_order(root)
    print("\nIn-order traversal:")
    in_order(root)
    print("\nPost-order traversal:")
    post_order(root)

Popularity 10/10 Helpfulness 2/10 Language python

Source: github.com

Tags: binary-tree bina

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Contributed on Jul 06 2022

Murad_9288

0 Answers Avg Quality 2/10

Tree: Postorder Traversal

Comment

def postOrder(root):
    if root:
        postOrder(root.left)
        postOrder(root.right)
        print(root, end = " ")

Popularity 10/10 Helpfulness 1/10 Language python

Source: Grepper

Tags: postorder post

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Contributed on Jun 18 2022

Grieving Gazelle

0 Answers Avg Quality 2/10

postorder traversal using stack

Comment

1. Right child of 1 exists. 
   Push 3 to stack. Push 1 to stack. Move to left child.
        Stack: 3, 1
2. Right child of 2 exists. 
   Push 5 to stack. Push 2 to stack. Move to left child.
        Stack: 3, 1, 5, 2
3. Right child of 4 doesn't exist. '
   Push 4 to stack. Move to left child.
        Stack: 3, 1, 5, 2, 4
4. Current node is NULL. 
   Pop 4 from stack. Right child of 4 doesn't exist. 
   Print 4. Set current node to NULL.
        Stack: 3, 1, 5, 2
5. Current node is NULL. 
    Pop 2 from stack. Since right child of 2 equals stack top element, 
    pop 5 from stack. Now push 2 to stack.     
    Move current node to right child of 2 i.e. 5
        Stack: 3, 1, 2
6. Right child of 5 doesn't exist. Push 5 to stack. Move to left child.
        Stack: 3, 1, 2, 5
7. Current node is NULL. Pop 5 from stack. Right child of 5 doesn't exist. 
   Print 5. Set current node to NULL.
        Stack: 3, 1, 2
8. Current node is NULL. Pop 2 from stack. 
   Right child of 2 is not equal to stack top element. 
   Print 2. Set current node to NULL.
        Stack: 3, 1
9. Current node is NULL. Pop 1 from stack. 
   Since right child of 1 equals stack top element, pop 3 from stack. 
   Now push 1 to stack. Move current node to right child of 1 i.e. 3
        Stack: 1
10. Repeat the same as above steps and Print 6, 7 and 3. 
    Pop 1 and Print 1.

Popularity 6/10 Helpfulness 1/10 Language whatever

Source: www.geeksforgeeks.org

Tags: postorder stack traversal trav

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Contributed on Sep 06 2024

Amjad Saeed

0 Answers Avg Quality 2/10

Inorder Traversal using Stack:

Comment

Inorder Traversal using Stack:
As we already know, recursion can also be implemented using stack. Here also we can use a stack to perform inorder traversal of a Binary Tree. Below is the algorithm for traversing a binary tree using stack.

Create an empty stack (say S).
Initialize the current node as root.
Push the current node to S and set current = current->left until current is NULL
If current is NULL and the stack is not empty then:
Pop the top item from the stack.
Print the popped item and set current = popped_item->right 
Go to step 3.

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Source: Grepper

Tags: inorder traversal using usin

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Contributed on Sep 06 2024

Amjad Saeed

0 Answers Avg Quality 2/10

Preorder traversal

Comment

Following is a simple stack based iterative process to print Preorder traversal. 

Create an empty stack nodeStack and push root node to stack. 
Do the following while nodeStack is not empty. 
Pop an item from the stack and print it. 
Push right child of a popped item to stack 
Push left child of a popped item to stack
The right child is pushed before the left child to make sure that the left subtree is processed first.

Popularity 5/10 Helpfulness 1/10 Language whatever

Source: Grepper

Tags: preorder traversal whatever

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Contributed on Sep 06 2024

Amjad Saeed

0 Answers Avg Quality 2/10

preorder traversal of binary tree

Comment

<> Preorder		Root -> Left -> Right 
<> Inorder		Left -> Root -> Right
<> Postorder	Left -> Right -> Root

Popularity 6/10 Helpfulness 1/10 Language whatever

Source: Grepper

Tags: binary-tree preorder traversal trav

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Contributed on Sep 06 2024

Amjad Saeed

0 Answers Avg Quality 2/10

Tree: Inorder Traversal

Comment

def inOrder(root):
    if root:
        inOrder(root.left)
        print(root, end = " ")
        inOrder(root.right)

Popularity 10/10 Helpfulness 1/10 Language python

Source: Grepper

Tags: inorder

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Contributed on Jun 18 2022

Grieving Gazelle

0 Answers Avg Quality 2/10

Tree traversal

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Tree: Inorder Traversal

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