Given the root
of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).
Example 1:
Input: root = [3,9,20,null,null,15,7] Output: [[3],[9,20],[15,7]]
Example 2:
Input: root = [1] Output: [[1]]
Example 3:
Input: root = [] Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 2000]
. -1000 <= Node.val <= 1000
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
if root is None:
return []
res = []
q = []
q.append(root)
while q:
size = len(q)
t = []
for _ in range(size):
node = q.pop(0)
if node.left is not None:
q.append(node.left)
if node.right is not None:
q.append(node.right)
t.append(node.val)
res.append(t)
return res
/**
* 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>> levelOrder(TreeNode root) {
if (root == null) return Collections.emptyList();
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
List<List<Integer>> res = new ArrayList<>();
while (!q.isEmpty()) {
int size = q.size();
List<Integer> t = new ArrayList<>();
while (size-- > 0) {
TreeNode node = q.poll();
t.add(node.val);
if (node.left != null) q.offer(node.left);
if (node.right != null) q.offer(node.right);
}
res.add(t);
}
return res;
}
}
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
if (!root) return {};
vector<vector<int>> res;
queue<TreeNode*> q{{root}};
while (!q.empty()) {
vector<int> oneLevel;
for (int i = q.size(); i > 0; --i) {
TreeNode* t = q.front();
q.pop();
oneLevel.push_back(t->val);
if (t->left) q.push(t->left);
if (t->right) q.push(t->right);
}
res.push_back(oneLevel);
}
return res;
}
};
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number[][]}
*/
var levelOrder = function (root) {
let result = [];
if (!root) return result;
let queue = [];
queue.push(root);
while (queue.length) {
let size = queue.length;
let levelItems = [];
while (size--) {
let node = queue.shift();
levelItems.push(node.val);
if (node.left) {
queue.push(node.left);
}
if (node.right) {
queue.push(node.right);
}
}
result.push(levelItems);
}
return result;
};
### **...**
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