LeetCode 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.

题意：已经二叉树的中序遍历和后序遍历，求二叉树的结构

思路：通过后序遍历中的最后一个元素，然后通过中序遍历知道左子树的包含元素和右子树的包含元素。根据左子树的中序遍历和后序遍历构造左子树，右子树的中序遍历和后序遍历构造右子树。

代码如下：

//二叉树的结构表示为
class TreeNode
{
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) {val = x;}
}
class Solution
{
    public TreeNode buildTree(int[] inorder, int[] postorder)
    {
        int len = postorder.length;
        if (0 == len) return null;

        TreeNode root = new TreeNode(postorder[len - 1]);
        int index = 0;
        for (int i = 0; i < inorder.length; i++)
        {
            if (inorder[i] == postorder[len - 1])
            {
                index = i;
                break;
            }
        }

        int[] leftinorder = new int[index];
        int[] leftpostorder = new int[index];

        int[] rightinorder = new int[len - index - 1];
        int[] rightpostorder = new int[len - index - 1];

        for (int i = 0; i < index; i++)
        {
            leftinorder[i] = inorder[i];
            leftpostorder[i] = postorder[i];
        }

        for (int i = 0; i < len - index - 1; i++)
        {

            rightinorder[i] = inorder[index + 1 + i];
            rightpostorder[i] = postorder[index + i];
        }

        root.left = buildTree(leftinorder, leftpostorder);
        root.right = buildTree(rightinorder, rightpostorder);

        return root;
    }
}