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.

思路：后序遍历中最后一个节点为根节点，在中序遍历找到根节点的位置，从而确定左子树和右子树的数量。在后序遍历和中序遍中划分了左子树、右子树节点的值之后，递归调用，分别创建左右子树。

/**
 * 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.length != postorder.length) {
            return null;
        }
        return myBuildTree(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);    
    }
    private int findPosition(int[] inorder, int start, int end, int key) {
        int i = 0;
        for(i = start; i <= end; i++) {
            if (inorder[i] == key) {
                return i;
            }
        }
        return -1;
    }
    
    private TreeNode myBuildTree(int[] inorder, int instart, int inend, int[] postorder, int pstart, int pend) {
        if (instart > inend ) {
            return null;
        }
        TreeNode root = new TreeNode(postorder[pend]);
        int position = findPosition(inorder, instart, inend, postorder[pend]);
        if (position == -1) {
            return null;
        }
        root.left = myBuildTree(inorder, instart, position - 1, postorder, pstart, pstart + position - instart - 1);
        root.right = myBuildTree(inorder, position + 1, inend, postorder, pstart + position - instart, pend - 1);
        return root;
    }
}