# Question

Formatted question description: https://leetcode.ca/all/366.html

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

# Algorithm

Each node is separated from the left child node and the right child node to get two depths,

Since the condition of becoming a leaf node is that the left and right child nodes are empty, we take the larger value of the left and right child nodes plus 1 as the depth value of the current node.

Knowing the depth value, you can add the node value to the correct position in the result.

# Code

• import java.util.ArrayList;
import java.util.List;

public class Find_Leaves_of_Binary_Tree {

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; // @note: +1==0, mapping to list index
}

int left = helper(list, root.left);
int right = helper(list, root.right);
int currentDepthFromBottom = 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() <= currentDepthFromBottom) {
}

return currentDepthFromBottom;
}
}

############

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public List<List<Integer>> findLeaves(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
TreeNode prev = new TreeNode(0, root, null);
while (prev.left != null) {
List<Integer> t = new ArrayList<>();
dfs(prev.left, prev, t);
}
return res;
}

private void dfs(TreeNode root, TreeNode prev, List<Integer> t) {
if (root == null) {
return;
}
if (root.left == null && root.right == null) {
if (prev.left == root) {
prev.left = null;
} else {
prev.right = null;
}
}
dfs(root.left, root, t);
dfs(root.right, root, t);
}
}

• // OJ: https://leetcode.com/problems/find-leaves-of-binary-tree/
// Time: O(N)
// Space: O(H)
class Solution {
private:
bool dfs(TreeNode *root, vector<int> &v) {
if (!root) return true;
if (!root->left && !root->right) {
v.push_back(root->val);
return true;
}
if (dfs(root->left, v)) root->left = NULL;
if (dfs(root->right, v)) root->right = NULL;
return false;
}
vector<int> removeLeaves(TreeNode *root) {
vector<int> v;
dfs(root, v);
return v;
}
public:
vector<vector<int>> findLeaves(TreeNode* root) {
if (!root) return {};
vector<vector<int>> ans;
while (root->left || root->right) {
ans.push_back(removeLeaves(root));
}
ans.push_back({ root->val });
return ans;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def findLeaves(self, root: TreeNode) -> List[List[int]]:
def dfs(root, prev, t):
if root is None:
return
if root.left is None and root.right is None:
t.append(root.val)
if prev.left == root:
prev.left = None
else:
prev.right = None
dfs(root.left, root, t)
dfs(root.right, root, t)

res = []
prev = TreeNode(left=root)
while prev.left:
t = []
dfs(prev.left, prev, t)
res.append(t)
return res

############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
import collections

class Solution(object):
def findLeaves(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""

def helper(p, res):
if not p:
return 0
left = helper(p.left, res)
right = helper(p.right, res)
depth = max(left, right) + 1
res[depth].append(p.val)
return depth

ans = []
res = collections.defaultdict(list)
helper(root, res)
for i in range(1, len(res) + 1):
ans.append(res[i])
return ans


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func findLeaves(root *TreeNode) [][]int {
prev := &TreeNode{
Val:   0,
Left:  root,
Right: nil,
}
var res [][]int
for prev.Left != nil {
var t []int
dfs(prev.Left, prev, &t)
res = append(res, t)
}
return res
}

func dfs(root, prev *TreeNode, t *[]int) {
if root == nil {
return
}
if root.Left == nil && root.Right == nil {
*t = append(*t, root.Val)
if prev.Left == root {
prev.Left = nil
} else {
prev.Right = nil
}
}
dfs(root.Left, root, t)
dfs(root.Right, root, t)
}