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102. Binary Tree Level Order Traversal.java
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102. Binary Tree Level Order Traversal.java
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102. Binary Tree Level Order Traversal
Solution 1: BFS
Time: O(n)
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) return res;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
List<Integer> tmp = new ArrayList<>();
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
tmp.add(node.val);
if (node.left != null) queue.offer(node.left);
if (node.right != null) queue.offer(node.right);
}
res.add(tmp);
}
return res;
}
Solution 2: DFS
Time: O(n)
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
dfs(res, root, 0);
return res;
}
private void dfs(List<List<Integer>> res, TreeNode root, int level) {
if (root == null) return;
if (level == res.size()) res.add(new ArrayList<>());
res.get(level).add(root.val);
dfs(res, root.left, level + 1);
dfs(res, root.right, level + 1);
}