# 106 Construct Binary Tree from Inorder and Postorder Traversal – Medium

### Problem:

Given inorder and postorder traversal of a tree, construct the binary tree.

Note: You may assume that duplicates do not exist in the tree.

### Thoughts:

This is silimar to Construct Binary Tree from Preorder and Inorder Traversal problem.

The only difference is that instead of finding the root from the first element, finding it at the last element in the postorder traversal.

### Solutions:

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {
if (postorder == null || inorder == null || postorder.length == 0 || inorder.length == 0 || postorder.length != inorder.length) {
return null;
}
HashMap<Integer, Integer> inmap = new HashMap<Integer, Integer>();
for (int i = 0; i < inorder.length; i ++) {
inmap.put(inorder[i], i);
}
return gt(postorder, inorder, 0, postorder.length - 1, 0, inorder.length - 1, inmap);
}
private TreeNode gt(int[] post, int[] in, int postl, int postr, int inl, int inr, HashMap<Integer, Integer> inmap) {
if (postl > postr) {
return null;
}
TreeNode root = new TreeNode(post[postr]);
int rootIndex = inmap.get(post[postr]);
int leftnum = rootIndex - inl;
TreeNode left = gt(post, in, postl, postl + leftnum -1, inl, rootIndex - 1, inmap);
TreeNode right = gt(post, in,postl + leftnum, postr - 1, rootIndex + 1, inr, inmap);
root.left = left;
root.right = right;
return root;
}
}
``````