106. Construct Binary Tree from Inorder and Postorder Traversal

public TreeNode buildTree(int[] inorder, int[] postorder) {
    if(inorder==null || postorder==null || inorder.length==0 || postorder.length==0)
    {
        return null;
    }
    HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
    for(int i=0;i<inorder.length;i++)
    {
        map.put(inorder[i],i);
    }
    return helper(inorder,postorder,0,inorder.length-1, 0, postorder.length-1,map);
}
private TreeNode helper(int[] inorder, int[] postorder, int inL, int inR, int postL, int postR,
 HashMap<Integer, Integer> map) {
    if(inL>inR || postL>postR)   //构造树的递归出口, 
        return null;
    TreeNode root = new TreeNode(postorder[postR]);
    int index = map.get(root.val);
    root.left = helper(inorder,postorder,inL,index-1,postL,postL+index-inL-1,map);
    root.right = helper(inorder,postorder,index+1,inR, postL+index-inL, postR-1,map);
    return root;
}

做个test case 看看起始点和终止点的位置, 构造树都不包括遍历过得