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());
    }
}