https://leetcode.com/problems/binary-tree-level-order-traversal/
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree [3,9,20,null,null,15,7],
3
/
9 20
/
15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
- code
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
LinkedList<TreeNode> level = new LinkedList<>();
if (root != null) level.offer(root);
while (!level.isEmpty()){
res.add(level.stream().map(n -> n.val).collect(Collectors.toList()));
int levelSize = level.size();
for (int i = 0; i < levelSize; i++){
TreeNode cur = level.pollFirst();
if (cur.left != null) level.offerLast(cur.left);
if (cur.right != null) level.offerLast(cur.right);
}
}
return res;
}
}
- code
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
if not root: return None
res = []
level = [root]
while level:
res.append([node.val for node in level])
level = [i for node in level for i in (node.left, node.right) if i]
return res
- c0
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
if not root:
return []
def traversal(root, level, res):
if not root:
return
if len(res) <= level:
res.append([])
res[level].append(root.val)
traversal(root.left, level+1, res)
traversal(root.right, level+1, res)
res = [[]]
traversal(root,0,res)
return res