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Postorder Traversal

Hassan answered on January 30, 2021 Popularity 10/10 Helpfulness 1/10

answer Postorder Traversal

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 10/10 Helpfulness 1/10 Language python

Source: Grepper

Tags: postorder python traversal

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Contributed on Jul 24 2023

Hassan

0 Answers Avg Quality 2/10

Closely Related Answers

inorder preorder postorder

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preorder: parent => left => right
inorder: left => parent => right
postorder: left => right => parent

Popularity 10/10 Helpfulness 4/10 Language whatever

Source: Grepper

Tags: inorder

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Contributed on Jan 30 2021

QuietHumility

0 Answers Avg Quality 2/10

inorder preorder postorder

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preorder: +ab
inorder: a+b
postorder: ab+
# Abhay Tiwari IITP

Popularity 9/10 Helpfulness 4/10 Language whatever

Source: Grepper

Tags: inorder

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Contributed on Aug 13 2023

TAbhay

0 Answers Avg Quality 2/10

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

inorder preorder postorder

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Hint to Remember: 
  IN means root is in middle: inorder: left => root => right 
  PRE means root is at front: preorder: root => left => right
  POST means root is at the end: postorder: left => right => root
Always left comes first than right

Popularity 10/10 Helpfulness 3/10 Language whatever

Source: Grepper

Tags: inorder

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Contributed on Sep 08 2022

Prajun Trital

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

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

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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 10/10 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

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 7/10 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:

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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.

Popularity 7/10 Helpfulness 1/10 Language whatever

Source: Grepper

Tags: inorder traversal tr

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Contributed on Sep 06 2024

Amjad Saeed

0 Answers Avg Quality 2/10

Postorder Traversal

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