Tree: Postorder Traversal


Problem Statement :


Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values.

Input Format

Our test code passes the root node of a binary tree to the postorder function.

Constraints

1 <=  Nodes in the tree   <= 500

Output Format

Print the tree's postorder traversal as a single line of space-separated values.

Sample Input

     1
      \
       2
        \
         5
        /  \
       3    6
        \
         4
Sample Output

4 3 6 5 2 1



Solution :



title-img


                            Solution in C :

In Java :


void Postorder(Node root) {
    if (root == null) { return; }
    Postorder(root.left);
    Postorder(root.right);
    System.out.print(root.data + " ");
}





In C++ :


/* you only have to complete the function given below.  
Node is defined as  

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

*/


void Postorder(node *root) {
    if (!root) return;
    Postorder(root->left);
    Postorder(root->right);
    printf("%d ", root->data);
}



In C :



/* you only have to complete the function given below.  
node is defined as  

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

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

}




In python3 :



"""
Node is defined as
self.left (the left child of the node)
self.right (the right child of the node)
self.data (the value of the node)
"""

def _preOrder(root, acc):
    if root:
        _preOrder(root.left, acc)
        _preOrder(root.right, acc)
        acc.append(root.data)

def postOrder(root):
    acc = []
    _preOrder(root, acc)
    print(" ".join(map(str, acc)))
    #Write your code here
                        








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