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:

Input: [1,2,3,4,5]
  
          1
         / \
        2   3
       / \     
      4   5    

Output: [[4,5,3],[2],[1]]

Solution： Bottom - up

1. 根据题目分析，可以发现，所有叶子节点如[4,5,3]，其height为1； [2]height为2； [1]height为3.
2. 所以可以采用bottom-up的方式，找到所有节点的height，把height相同的节点归成一类，就是答案。
3. 归类可以用HashMap数据结构： key: height, value: List of Node

// hashmap to store, height and all nodes with this height
Map<Integer, List<Integer>> tracker = new HashMap<> ();

Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> findLeaves(TreeNode root) {
        List<List<Integer>> result = new ArrayList<> ();
        if (root == null) {
            return result;
        }
        
        // hashmap to store, height and all nodes with this height
        Map<Integer, List<Integer>> tracker = new HashMap<> ();
        
        // recurse to popular the hashmap
        getHeights (root, tracker);
        
        for (Map.Entry <Integer, List<Integer>> entrySet : tracker.entrySet ()) {
            result.add (entrySet.getValue ());
        }
        
        return result;
    }
    
    private int getHeights (TreeNode root, Map<Integer, List<Integer>> tracker) {
        if (root == null) {
            return 0;
        }
        
        int leftHeight = getHeights (root.left, tracker);
        int rightHeight = getHeights (root.right, tracker);
        
        int currentHeight = Math.max (leftHeight, rightHeight) + 1;
        
        List<Integer> nodes = tracker.getOrDefault (currentHeight, new ArrayList<Integer> ());
        nodes.add (root.val);
        tracker.put (currentHeight, nodes);

        return currentHeight;
    }
}