Given the root of a binary tree, collect a tree's nodes as if you were doing this:
- Collect all the leaf nodes.
- Remove all the leaf nodes.
- Repeat until the tree is empty.
Example 1:
Input: root = [1,2,3,4,5] Output: [[4,5,3],[2],[1]] Explanation: [[3,5,4],[2],[1]] and [[3,4,5],[2],[1]] are also considered correct answers since per each level it does not matter the order on which elements are returned.
Example 2:
Input: root = [1] Output: [[1]]
Constraints:
- The number of nodes in the tree is in the range
[1, 100]. -100 <= Node.val <= 100
题目给定一棵二叉树，要求把所有叶节点值放入一个数组中，然后把所有叶节点都删除，这样树中就又有一些节点变成了新的叶节点，重复同样的动作就像剥洋葱一样一层一层地把叶节点去掉，直到最后只剩一个根节点。
因此一个节点在第几层被剥离决定了它在结果中的位置，而一个节点是在第几层被剥离又取决于它的左右子树的最大高度，根据题意要剥离一个节点肯定要等到其所有子节点都被剥离了，因此左右子树最大高度的更大者加1就是那个节点的层数。
class Solution:
def findLeaves(self, root: Optional[TreeNode]) -> List[List[int]]:
if not root:
return []
res = []
def helper(node):
nonlocal res
if not node:
return 0
if not node.left and not node.right:
if len(res) == 0:
res.append([])
res[0].append(node.val)
return 1
ld = helper(node.left)
rd = helper(node.right)
d = max(ld, rd)
if len(res) == d:
res.append([])
res[d].append(node.val)
return d + 1
helper(root)
return res
