Postorder TraversalThe postorder traversal is one of the traversing techniques used for visiting the node in the tree. It follows the principle LRN (Leftrightnode). The following are the steps used for the postorder traversal:
Pseudocode of postorder traversal. In the above pseudocode, we have defined a postorder function in which we added the logic of postorder traversal. As we already know that in postorder traversal, we first traverse the left subtree, then the right subtree and finally visit the root node. In the above code, we first have checked whether the value of root is NULL or not. If the value of root is NULL, then the control comes out of the function using the return statement. The 'if' condition is not satisfied, then the else part will be executed. Implementation of Postorder traversal in C. Output Iterative Postorder traversalThe above method that we have used for the Postorder traversal was the Recursion method. As we know that the recursion relies on the system stack, and if we want to convert the recursion into an iterative method, we need to use the external stack. There are two ways of iterative postorder traversal:
Iterative Postorder traversal using two stacks We can also implement the Postorder traversal using two stacks. The idea behind this is to obtain the reversed postorder elements in the stack and then we pop the element one by one from the stack as the stack follows the LIFO principle. The question arises that how can we obtain the reversed order of the postorder elements. The solution to this problem can be obtained by using two stacks. The second stack is used to get the reversed order of the postorder elements. The following are the steps used to implement postorder using two stacks:
Let's understand the postorder traversal using two stacks through an example. Consider the below tree for the Postorder traversal. Steps to be followed: Step 1: First, we push 10 into stack1. Stack1: 10 Stack2: empty Step 2: Now we pop element 10 from stack1 and push it into stack2. We will push the left and right child of element 10 into stack1. Stack1: 20, 30 Stack2: 10 Step 3: Now, we pop element 30 from stack1 and push it into stack2. We will push the left and right child of element 30 into stack1: Stack 1: 20, 60, 70 Stack 2: 10, 30 Step 4: Now we pop the element 70 from the stack1 and push it into the stack2. We will push the left and right child of element 70 into the stack1. Since there is no left and right child of element 70, so we will not push the element into Stack1. Stack1: 20, 60 Stack2: 10, 30, 70 Step 5: Now we pop the element 60 from the Stack1 and push it into Stack2. Since there is no left and right of element 60, so we will not push any element into Stack1. Stack1: 20 Stack2: 10, 30, 70, 60 Step 6: Now, we pop the element 20 from Stack1 and push it into Stack2. We will push the left and right child of element 20 into the Stack1. Stack1: 40, 50 Stack2: 10, 30, 70, 60, 20 Step 7: Now, we pop the element 50 from Stack1 and push it into Stack2. Stack1: 40 Stack2: 10, 30, 70, 60, 20, 50 Step 8: Now, we pop element 40 from the Stack1 and push it into Stack2. Stack1: Empty Stack2: 10, 30, 70, 60, 20, 50, 40 Since there is no element left in the Stack1 to be popped out and all the elements in Stack2 are in postorder fashion, so we will print them. Algorithm of iterative postorder traversal (two stacks) Iterative Postorder Traversal using one stack In order to implement the Postorder traversal using one stack, first we need to move down till we reach the leftmost node. While moving down to the tree, we need to push the root and then right child of the root into the stack. When we reach the leftmost node having no right child, we need to print the node. If the leftmost node has a right child, we have to update the root so that the right child can be traversed before. We have to perform three operations:
Let's understand through an example. Step 1: The right child of node 1 exists. Push node 3 to the stack and then 1 to the stack. Stack: 3, 1 Step 2: The right child of node 2 exists. Push node 5 to the stack and then 2 to the stack. Stack: 3, 1, 5, 2 Step 3: The node 4 has no right child, so it gets printed and the current node is set to Null. Stack: 3, 1, 5, 2 Output: 4 Step 4: Now, peek operation will be performed on the node 2. Since the right child of node 2 is topmost element in the stack, i.e., 5. So, pop the element 5 from the stack. Stack: 3, 1, 2 Step 5: The right child of node 2 is 5, so we push 5 into the stack. Stack: 3, 1, 2, 5 Step 6: Since node 5 does not have a right child so pop 5 from the stack and print 5. The current node is now set to NULL. Stack: 3, 1, 2 Output: 4, 5 Step 7: The current node is NULL. Since the right child of node 2 is not the topmost element in the stack, so we will pop element 2 from the stack and print 2. The current node is now set to NULL. Stack: 3, 1 Output: 4, 5, 2 Step 8: The current node is NULL. Since the right child of node 1 is 3 so we will perform the peek operation on node 1 and pop 3 from the stack. Stack: 1 Step 9: The right child of 1 is 3 so we will push element 3 in the stack. Stack: 1, 3 Step 10: The node 3 has a left child, i.e., 6, so it gets printed and the right child of node 3, i.e., 7 is pushed into the stack. Stack: 1, 3, 7 Output: 4, 5, 2, 6 Step 11: Since there is no right child of node 7, so pop 7 from the stack. Stack: 1, 3 Output: 4, 5, 2, 6, 7 Step 12: Since the right child of node 3 is not the topmost element in the stack, so pop 3 from the stack. Stack: 1 Output: 4, 5, 2, 6, 7, 3 Step 12: Since the right child of node 1 is not the topmost element in the stack, so Pop 1 from the stack. Stack: empty Output: 4, 5, 2, 6, 7, 3, 1 The above output generated is the Postorder traversal. Algorithm of iterative postorder traversal (one stack)
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