leetCode-Construct Binary Tree from Inorder and Postorder Traversal

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

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

Solution:

//中序序列和后序序列构造二叉树，与前序序列和中序序列构建二叉树异曲同工，都是利用前序或者后序序列找出根节点(前序序列首端，后序序列末端为根节点),然后利用根节点在中序序列中找出左右子树的个数，然后在分割前序序列或者后序序列，之后递归构建树.如果有不明白的地方可以看看这篇博文:http://blog.csdn.net/zhangjiajie023/article/details/7912757/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */class Solution {    public TreeNode buildTree(int[] inorder, int[] postorder) {        Map<Integer,Integer>map = new HashMap<Integer,Integer>();        for(int i = 0;i < inorder.length;i++){            map.put(inorder[i],i);        }        return inpoTree(inorder,0,inorder.length - 1,postorder,0,postorder.length - 1,map);    }    TreeNode inpoTree(int[] inorder,int instart,int inend,int[] postorder,int poststart,int postend,Map<Integer,Integer>map){        if(instart > inend || poststart > postend){            return null;        }        TreeNode node = new TreeNode(postorder[postend]);        int index = map.get(postorder[postend]);        int numsleft = index - instart;        int numsright = inend - index;        node.left = inpoTree(inorder,instart,index - 1,postorder,poststart,poststart + numsleft - 1,map);        node.right = inpoTree(inorder,index + 1,inend,postorder,poststart + numsleft,postend - 1,map);        return node;    }}