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;
    }
}