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 :
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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