106.Construct Binary Tree from Inorder and Postorder Traversal.md

Given inorder and postorder traversal of a tree, construct the binary tree.
Note: You may assume that duplicates do not exist in the tree.

题目大意：根据中序和后序遍历，构造二叉树

题目难度：Medium

import java.util.*;
/**
 * Created by gzdaijie on 16/6/7
 */
public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        return buildTreeHelper(inorder, 0, postorder, 0, inorder.length);
    }

    private TreeNode buildTreeHelper(int[] inorder, int index1, int[] postorder, int index2, int len) {
        if (len == 0) return null;

        int num = postorder[index2 + len - 1];
        int count = 0;
        TreeNode root = new TreeNode(num);

        for (int i = 0; i < len; i++) {
            if (inorder[index1 + i] == num) break;
            count++;
        }
        root.left = buildTreeHelper(inorder, index1, postorder, index2, count);
        root.right = buildTreeHelper(inorder, index1 + count + 1, postorder, index2 + count, len - count - 1);
        return root;
    }
}