FB面经 Prepare: Even Tree

You are given a tree (a simple connected graph with no cycles). The tree has  nodes numbered from  to  and is rooted at node .

Find the maximum number of edges you can remove from the tree to get a forest such that each connected component of the forest contains an even number of vertices.

        o
  /    |   |   \
o     o   o    o
|
o

比如可以删掉一个边变成：
         o
  x    |   |   \
o     o   o    o
|
o
结果里有两个tree，分别有2个和四个node，符合条件，这就是答案，因为再删就不符合条件了
return是一个list，里面是所有新生成的tree的root

我觉得题中应该再加上一个条件，就是guarantee是能够分割的，不然没法做. 如果总node总数是奇数的话， 怎么删都没法保证所有的子树是even number，所以这题的前提是node总数为偶数？

网上看到别人的很好的解法：

特别是用iterator.next()以后用iterator.remove()

 public class TreeNode{
     int val;
     List<TreeNode> subtree;
     public TreeNode(int val){
         this.val = val;.
         subtree = new ArrayList<>();
     }

     public void addChild(TreeNode child){
         subtree.add(child);
     }
 }

 public class BreakTree {
     public List<TreeNode> breakTree(TreeNode root){
         List<TreeNode> result = new ArrayList<>();
         countAndBreak(result, root);
         return result;
 }

     private int countAndBreak(List<TreeNode> result, TreeNode root){
         if (root == null){
             return 0;
         }

         Iterator<TreeNode> iter = root.subtree.iterator();
         while (iter.hasNext()){
             int childCount = countAndBreak(result, iter.next());
             if (childCount == 0){
                 iter.remove();
             } else{
                 count += childCount;
             }
         }
         if (count % 2 == 0){
             result.add(root);
             return 0;
         } else{
             return count;
         }
     }

     public static void main(String[] args){
         TreeNode root = new TreeNode(0);

         TreeNode firstChild = new TreeNode(1);
         firstChild.addChild(new TreeNode(2));
         root.addChild(firstChild);

         root.addChild(new TreeNode(3));
         root.addChild(new TreeNode(4));
         root.addChild(new TreeNode(5));

         BreakTree soln = new BreakTree();
         List<TreeNode> result = soln.breakTree(root);
         System.out.println(result.size());
     }
 }