# Construct a binary tree from given Inorder and Postorder Traversal

Objective: Given a inorder and postorder traversal, write an algorithm to construct a binary tree from that. This problem was asked in the Microsoft coding competition.

Input: Inorder and postorder traversals

Similar Problems: Make a Binary Tree from Given Inorder and Preorder Traveral.

Appraoch:

int[] inOrder = { 4, 2, 5, 1, 6, 3, 7 };

int[] postOrder = { 4, 5, 2, 6, 7, 3, 1 };.

• Last element in the postorder [] will be the root of the tree, here it is 1.
• Now the search ele­ment 1 in inorder[], say you find it at posi­tion i, once you find it, make note of ele­ments which are left to i (this will con­struct the left­sub­tree) and ele­ments which are right to i ( this will con­struct the rightSubtree).
• Suppose in previous step, there are X number of elements which are left of ‘i’ (which will construct the leftsubtree), take first X elements from the postorder[] traversal, this will be the post order traversal for elements which are left to i. similarly if there are Y number of elements which are right of ‘i’ (which will construct the rightsubtree), take next Y elements, after X elements from the postorder[] traversal, this will be the post order traversal for elements which are right to i
• From previous two steps construct the left and right subtree and link it to root.left and root.right respectively.
• See the picture for better explanation.

Complete Code:

 import java.util.LinkedList; import java.util.Queue; public class InorderPostOrderToTree { public static int pIndex = 0; public Node makeBTree(int[] inOrder, int[] postOrder, int iStart, int iEnd, int postStart, int postEnd) { if (iStart > iEnd || postEnd > postEnd) { return null; } int rootValue = postOrder[postEnd]; Node root = new Node(rootValue); pIndex++; if (iStart == iEnd) { return root; } int index = getInorderIndex(inOrder, iStart, iEnd, root.data); root.left = makeBTree(inOrder, postOrder, iStart, index – 1, postStart, postStart + index – (iStart + 1)); root.right = makeBTree(inOrder, postOrder, index + 1, iEnd, postStart + index – iStart, postEnd – 1); // } return root; } public int getInorderIndex(int[] inOrder, int start, int end, int data) { for (int i = start; i <= end; i++) { if (inOrder[i] == data) { return i; } } return –1; } public void printINORDER(Node root) { if (root != null) { printINORDER(root.left); System.out.print(" " + root.data); printINORDER(root.right); } } public static void main(String[] args) throws java.lang.Exception { int[] inOrder = { 4, 2, 5, 1, 6, 3, 7 }; int[] postOrder = { 4, 5, 2, 6, 7, 3, 1 }; InorderPostOrderToTree i = new InorderPostOrderToTree(); Node x = i.makeBTree(inOrder, postOrder, 0, inOrder.length – 1, 0, postOrder.length – 1); System.out.println("inorder traversal of constructed tree : "); i.printINORDER(x); } } class Node { int data; Node left; Node right; public Node(int data) { this.data = data; left = null; right = null; } }

Output:

```inorder traversal of constructed tree :
4  2  5  1  6  3  7
```

### 1 thought on “Construct a binary tree from given Inorder and Postorder Traversal”

1. good work!is the worst case complexity O(n^2)?