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Tree: Inorder Traversal

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Grieving Gazelle answered on May 27, 2022 Popularity 10/10 Helpfulness 1/10

answer Tree: Inorder Traversal

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

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 python traversal

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Contributed on Jun 18 2022

Grieving Gazelle

0 Answers Avg Quality 2/10

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preorder traversal

Comments(1)

<> 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

Tree traversal

Comment

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

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Source: Grepper

Tags: inorder

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Contributed on Aug 03 2022

ap_Cooperative_dev

0 Answers Avg Quality 2/10

inorder traversal of tree

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//*******this is a C program for implementation and searching in A BT*******
#include<stdlib.h>
#include <stdio.h>

struct BinaryTree{
    int data;
    struct BinaryTree*right;
    struct BinaryTree*left;
};

struct BinaryTree*createnode(int val){
    struct BinaryTree*root=(struct BinaryTree*)malloc(sizeof(struct BinaryTree));
    root->data=val;
    root->left=NULL;
    root->right=NULL;
    
}

void inorder(struct BinaryTree*root){
    if(root==NULL){
    return;
    }
    else {inorder(root->left);
    printf("%d->",root->data);
    inorder(root->right);
}

}
 
void preorder(struct BinaryTree*root){
    if(root==NULL){
        return;
    }
    else {
        printf("%d->",root->data);
        preorder(root->left);
        preorder(root->right);
    }
}

void postorder(struct BinaryTree*root){
    if(root==NULL){
        return;
    }
    else{
        postorder(root->left);
        postorder(root->right);
        printf("%d->",root->data);
    }
}
int main()
{
    printf("Lets grow the tree\n");
    struct BinaryTree*root=createnode(1);
    root->left=createnode(2);
    root->right=createnode(3);
    root->left->left=createnode(4);
    root->left->right=createnode(5);
    
    printf("tree has grown up\n");
    
    printf("Inorder traversal ");
    inorder(root);printf("NULL");

    printf("\npreorder traversal ");
    preorder(root);printf("NULL");
    
    printf("\nPostorder  traversal");
    postorder(root);printf("NULL");
    
    return 0 ;
}

Popularity 10/10 Helpfulness 2/10 Language c

Source: Grepper

Tags: c

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Contributed on Aug 17 2022

Unusual Unicorn

0 Answers Avg Quality 2/10

Inorder traversal Algorithm for binary tree

Comment

Until all nodes are traversed −
Step 1 − Recursively traverse left subtree.
Step 2 − Visit root node.
Step 3 − Recursively traverse right subtree.

Popularity 10/10 Helpfulness 2/10 Language whatever

Source: www.tutorialspoint.com

Tags: algorithm algo

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Contributed on Sep 22 2022

Taiwo Adedotun

0 Answers Avg Quality 2/10

traversal tree

Comment

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

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

Comment

/*
	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 1/10 Language java

Source: Grepper

Tags: java

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Contributed on May 27 2022

Wissam

0 Answers Avg Quality 2/10

Postorder Traversal

Comment

# Assuming valid Binary Tree Node
def post_order(node: BinaryTreeNode):
    if node.left:
        post_order(node.left)
    if node.right:
        post_order(node.right)
    print(node.val)

Popularity 9/10 Helpfulness 1/10 Language python

Source: Grepper

Tags: postorder post

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Contributed on Jul 24 2023

Hassan

0 Answers Avg Quality 2/10

Postorder traversal

Comment

class Solution {
    List<Integer> list = new ArrayList<>();
    public List<Integer> postorderTraversal(TreeNode root) {
        dfs(root);
        return list;
        
    }
     public void dfs(TreeNode root)
    {
        if(root!=null)
        {
            
            dfs(root.left);
            
            dfs(root.right);
            list.add(root.val);
        }
    }
}

Popularity 9/10 Helpfulness 1/10 Language whatever

Source: Grepper

Tags: postorder post

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Contributed on Jun 22 2023

Zealous Zebra

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

preorder traversal of binary tree c++

Comment

#include <bits/stdc++.h>

using namespace std;

struct node {
  int data;
  struct node * left, * right;
};

vector < int > preOrderTrav(node * curr) {
  vector < int > preOrder;
  if (curr == NULL)
    return preOrder;

  stack < node * > s;
  s.push(curr);

  while (!s.empty()) {
    node * topNode = s.top();
    preOrder.push_back(topNode -> data);
    s.pop();
    if (topNode -> right != NULL)
      s.push(topNode -> right);
    if (topNode -> left != NULL)
      s.push(topNode -> left);
  }
  return preOrder;

}

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);
}

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 -> left -> right -> left = newNode(8);
  root -> right -> left = newNode(6);
  root -> right -> right = newNode(7);
  root -> right -> right -> left = newNode(9);
  root -> right -> right -> right = newNode(10);

  vector < int > preOrder;
  preOrder = preOrderTrav(root);

  cout << "The preOrder Traversal is : ";
  for (int i = 0; i < preOrder.size(); i++) {
    cout << preOrder[i] << " ";
  }
  return 0;
}

Popularity 8/10 Helpfulness 1/10 Language cpp

Source: takeuforward.org

Tags: binary-tree bina

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Contributed on Dec 10 2023

Hitesh Prajapati

0 Answers Avg Quality 2/10

Tree: Inorder Traversal

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Tree: Inorder Traversal

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