Construct Binary Tree from Inorder and Postorder Traversal

Construct Binary Tree from Inorder and Postorder Traversal

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

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

/**
 * 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 (inorder == null || postorder == null || inorder.length == 0 || postorder.length == 0)
            return null;
        
        int treeLength = inorder.length;
        
        return buildTreeHelper(inorder, 0, treeLength - 1, postorder, 0, treeLength - 1);
    }
    
    private TreeNode buildTreeHelper(int[] inorder, int inStart, int inEnd, int[] postorder, int postStart, int postEnd) {
        if (inStart > inEnd || postStart > postEnd)
            return null;
        
        int rootVal = postorder[postEnd];
        int rootIndex = 0;
        
        for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == rootVal) {
                rootIndex = i;
                break;
            }
        }
        
        TreeNode root = new TreeNode(rootVal);
        int leftLength = rootIndex - inStart;
        int rightLength = inEnd - rootIndex;
        
        root.left = buildTreeHelper(inorder, inStart, rootIndex - 1, postorder, postStart, postStart + leftLength - 1);
        root.right = buildTreeHelper(inorder, rootIndex + 1, inEnd, postorder, postStart + leftLength, postStart + leftLength + rightLength - 1);
        
        return root;
    }
}