linked list焦点问题，面经里很多，考虑相交不相交，有环无环 + Find Leaves of Binary Tree (Java)

break the loop at the last node which pointed to the entry.

Given a binary tree, collect a tree‘s nodes as if you were doing this: Collect and remove all leaves, repeat until the tree is empty.

Example:
Given binary tree

          1
         /         2   3
       / \
      4   5

Returns [4, 5, 3], [2], [1].

ollowup： 每个node还是有left，right指针，但是结构并非tree，而是graph怎么算， 考虑有/无circular dependency 两种情况；

方法是用hashmap 记录计算过的node level 和 一个Set 记录dependent node

public List<List<Integer>> findLeaves(TreeNode root) {
    List<List<Integer>> result = new ArrayList<List<Integer>>();
    helper(result, root);
    return result;
}

// traverse the tree bottom-up recursively
private int helper(List<List<Integer>> list, TreeNode root){
    if(root==null)
        return -1;

    int left = helper(list, root.left);
    int right = helper(list, root.right);
    int curr = Math.max(left, right)+1;

    // the first time this code is reached is when curr==0,
    //since the tree is bottom-up processed.
    if(list.size()<=curr){
        list.add(new ArrayList<Integer>());
    }

    list.get(curr).add(root.val);

    return curr;
}