> For the complete documentation index, see [llms.txt](https://maksimdan.gitbook.io/interview-practice-problems/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://maksimdan.gitbook.io/interview-practice-problems/leetcode_sessions/366-find-leaves-of-binary-tree.md).

# 366 Find Leaves of Binary Tree

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].
```

**Explanation:**

Removing the leaves \[4, 5, 3] would result in this tree:

```
          1
         / 
        2
```

Now removing the leaf \[2] would result in this tree:

```
          1
```

Now removing the leaf \[1] would result in the empty tree:

```
          []
```

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

**The Idea:** The height of any node is the longest path from the node to any leaf. Notice how we can use the height of the tree as a direct index to the solution. In other words, grouping and sorting the heights of the tree reveals the solution. Alternatively, we can map all common nodes to a particular height. Then we can iterate through all the heights in order to reveal the solution.

**Complexity:** O(n) time and space

```python
def findLeaves(self, root):
    """
    :type root: TreeNode
    :rtype: List[List[int]]
    """

    heights = collections.defaultdict(list)
    def dfs(root):
        if not root:
            return -1
        height = 1 + max(dfs(root.left), dfs(root.right))
        heights[height].append(root.val)
        return height

    if not root:
        return []

    dfs(root)
    return [heights[lvl] for lvl in range(0, len(heights))]
```


---

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