leetcode106~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.

根据中序遍历和后序遍历创建一棵二叉树。
思路与上题类似。
关键在于，上下限的选取。下面两种方法，不同在于上下限选取不一样。

public class BuildTree106 {    public TreeNode buildTree(int[] inorder,int[] postorder) {        int inLength = inorder.length;        int postLength = postorder.length;        return helper(inorder,0,inLength-1,postorder,0,postLength-1);    }    private TreeNode helper(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=0;i<=inend;i++) {            if(inorder[i]==rootval) {                rootIndex = i;                break;            }        }        int len = inend-rootIndex;        TreeNode root = new TreeNode(rootval);        root.left = helper(inorder,instart,rootIndex-1,postorder,poststart,postend-len-1);        root.right = helper(inorder,rootIndex+1,inend,postorder,postend-len,postend-1);        return root;    }    private TreeNode helper2(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=0;i<=inend;i++) {            if(inorder[i]==rootval) {                rootIndex = i;                break;            }        }        int len = rootIndex-instart;        TreeNode root = new TreeNode(rootval);        root.left = helper(inorder,instart,rootIndex-1,postorder,poststart,poststart+len-1);        root.right = helper(inorder,rootIndex+1,inend,postorder,poststart+len,postend-1);        return root;    }