divided-and-conquer-106. Construct Binary Tree from Inorder and Postorder Traversal

/** * 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) {        if(inorder == null || inorder.length == 0) return null;        //建立一个hash表，记录inorder数组的值->索引的映射        Map<Integer,Integer> map = new HashMap<Integer,Integer>();        for(int i = 0; i < inorder.length; i++){            map.put(inorder[i],i);        }        return helper(inorder,0,inorder.length-1,postorder,0,postorder.length-1,map);    }        public TreeNode helper(int[] inorder,int is,int ie,int[] postorder,int ps,int pe,Map<Integer,Integer> map){        if(is > ie || ps > pe) return null;        //取出根节点        TreeNode root = new TreeNode(postorder[pe]);        //获取当前根节点的值在中序数组的索引，并将数组分为两部分        int index = map.get(postorder[pe]);        //divided        TreeNode left = helper(inorder,is,index-1,postorder,ps,ps+index-is-1,map);        TreeNode right = helper(inorder,index+1,ie,postorder,ps+index-is,pe-1,map);        //conquer        root.left = left;        root.right = right;        return root;    }}