Prerequisites:- Morris Inorder Traversal, Tree Traversals (Inorder, Preorder and Postorder)

Given a Binary Tree, the task is to print the elements in post order using O(N) time complexity and constant space.

Input:   1 
       /   \
     2       3
    / \     / \
   4   5   6   7
  / \
 8   9
Output: 4 8 9 5 2 6 7 3 1

Input:   5 
       /   \
     7       3
    / \     / \
   4   11  13  9
  / \
 8   4
Output: 4 8 4 11 7 13 9 3 5 

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1: Using Morris Inorder Traversal

1. Create a dummy node and make the root as it’s left child.
2. Initialize current with dummy node.
3. While current is not NULL

• If current does not have a left chil traverse the right child, current = current->right
• Otherwise,

1. Find the right most child in the left subtree.
2. If right most child’s right child is NULL

• Make current as the right child of the right most node.
• Traverse the left child, current = current->left

1. Otherwise,

• Set the right most child’s right pointer to NULL.
• From current’s left child, traverse along the right children until the right most child and reverse the pointers.
• Traverse back from right most child to current’s left child node by reversing the pointers and printing the elements.

1. Traverse the right child, current = current->right

Below is the diagram showing the right most child in left subtree, pointing to it’s inorder successor.

Below is the diagram which highlights the path 1->2->5->9 and the way the nodes are processed and printed as per the above algorithm.

Below is the implementation of the above approach:

• Java

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// Java program to implement

// Post Order traversal

// of Binary Tree in O(N)

// time and O(1) space

// Definition of the

// binary tree

**class** TreeNode {

**public** **int** data;

**public** TreeNode left;

**public** TreeNode right;

**public** TreeNode(``**int** data)

{

**this**``.data = data;

}

**public** String toString()

{

**return** data + " "``;

}

}

**public** **class** PostOrder {

TreeNode root;

// Function to find Post Order

// Traversal Using Constant space

**void** postOrderConstantspace(TreeNode

root)

{

**if** (root == **null**``)

**return**``;

TreeNode current

= **new** TreeNode(-``1``),

pre = **null**``;

TreeNode prev = **null**``,

succ = **null**``,

temp = **null**``;

current.left = root;

**while** (current != **null**``) {

// Go to the right child

// if current does not

// have a left child

**if** (current.left == **null**``) {

current = current.right;

}

**else** {

// Traverse left child

pre = current.left;

// Find the right most child

// in the left subtree

**while** (pre.right != **null**

&& pre.right != current)

pre = pre.right;

**if** (pre.right == **null**``) {

// Make current as the right

// child of the right most node

pre.right = current;

// Traverse the left child

current = current.left;

}

**else** {

pre.right = **null**``;

succ = current;

current = current.left;

prev = **null**``;

// Traverse along the right

// subtree to the

// right-most child

**while** (current != **null**``) {

temp = current.right;

current.right = prev;

prev = current;

current = temp;

}

// Traverse back from

// right most child to

// current's left child node

**while** (prev != **null**``) {

System.out.print(prev);

temp = prev.right;

prev.right = current;

current = prev;

prev = temp;

}

current = succ;

current = current.right;

}

}

}

}

// Driver Code

**public** **static** **void** main(String[] args)

{

/* Constructed tree is as follows:-

1 

/     \

2       3

/ \     / \

4   5   6   7

/ \

8   9

*/

PostOrder tree = **new** PostOrder();

tree.root = **new** TreeNode(``1``);

tree.root.left = **new** TreeNode(``2``);

tree.root.right = **new** TreeNode(``3``);

tree.root.left.left = **new** TreeNode(``4``);

tree.root.left.right

= **new** TreeNode(``5``);

tree.root.right.left

= **new** TreeNode(``6``);

tree.root.right.right

= **new** TreeNode(``7``);

tree.root.left.right.left

= **new** TreeNode(``8``);

tree.root.left.right.right

= **new** TreeNode(``9``);

tree.postOrderConstantspace(

tree.root);

}

}

Output:

4 8 9 5 2 6 7 3 1

_Time Complexity: _O(N)

Auxiliary Space:_ O(1)_

Method 2: In method 1, we traverse a path, reverse references, print nodes as we restore the references by reversing them again. In method 2, instead of reversing paths and restoring the structure, we traverse to parent node from the current node using the current node’s left subtree. This could be faster depending on the tree structure, for example in a right-skewed tree.

The following algorithm and diagrams provide the details of the approach.

#data structures #recursion #tree #inorder traversal #interview-preparation #morris-traversal #postorder traversal