Description#
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
Solutions#
Solution 1#
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| # 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
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| /**
* 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);
res.add(t);
}
return res;
}
private void dfs(TreeNode root, TreeNode prev, List<Integer> t) {
if (root == null) {
return;
}
if (root.left == null && root.right == null) {
t.add(root.val);
if (prev.left == root) {
prev.left = null;
} else {
prev.right = null;
}
}
dfs(root.left, root, t);
dfs(root.right, root, t);
}
}
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| /**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> findLeaves(TreeNode* root) {
vector<vector<int>> res;
TreeNode* prev = new TreeNode(0, root, nullptr);
while (prev->left) {
vector<int> t;
dfs(prev->left, prev, t);
res.push_back(t);
}
return res;
}
void dfs(TreeNode* root, TreeNode* prev, vector<int>& t) {
if (!root) return;
if (!root->left && !root->right) {
t.push_back(root->val);
if (prev->left == root)
prev->left = nullptr;
else
prev->right = nullptr;
}
dfs(root->left, root, t);
dfs(root->right, root, t);
}
};
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| /**
* 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)
}
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