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

 

For this question we need to take bottom-up approach. The key is to find the height of each node. Here the definition of height is:
The height of a node is the number of edges from the node to the deepest leaf. 

I used a helper function to return the height of current node. According to the definition, the height of leaf is 0. h(node) = 1 + max(h(node.left), h(node.right)).
The height of a node is also the its index in the result list (res). For example, leaves, whose heights are 0, are stored in res[0]. Once we find the height of a node, we can put it directly into the result.

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;
}