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102. Binary Tree Level Order Traversal
这个题其实很简单，只要保持需要读取值的那一层的节点就可以了。
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] ]
思路分析
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一刷
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二刷
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三刷
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/**
* Runtime: 3 ms, faster than 5.15% of Java online submissions for Binary Tree Level Order Traversal.
*
* Memory Usage: 40.2 MB, less than 5.33% of Java online submissions for Binary Tree Level Order Traversal.
*/
public List<List<Integer>> levelOrder(TreeNode root) {
if (Objects.isNull(root)) {
return Collections.emptyList();
}
List<List<Integer>> result = new ArrayList<>();
List<TreeNode> level = new ArrayList<>();
level.add(root);
while (!level.isEmpty()) {
List<TreeNode> temp = new ArrayList<>();
List<Integer> values = new ArrayList<>(level.size());
for (TreeNode node : level) {
values.add(node.val);
if (Objects.nonNull(node.left)) {
temp.add(node.left);
}
if (Objects.nonNull(node.right)) {
temp.add(node.right);
}
}
result.add(values);
level = temp;
}
return result;
}
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/**
* Runtime: 3 ms, faster than 5.15% of Java online submissions for Binary Tree Level Order Traversal.
* <p>
* Memory Usage: 40.2 MB, less than 5.33% of Java online submissions for Binary Tree Level Order Traversal.
*
* @author D瓜哥 · https://www.diguage.com
* @since 2024-06-25 11:27
*/
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
if (root == null) {
return result;
}
Queue<TreeNode> topQueue = new LinkedList<>();
topQueue.add(root);
while (!topQueue.isEmpty()) {
List<Integer> level = new ArrayList<>();
Queue<TreeNode> bottomQueue = new LinkedList<>();
while (!topQueue.isEmpty()) {
TreeNode top = topQueue.poll();
level.add(top.val);
if (top.left != null) {
bottomQueue.add(top.left);
}
if (top.right != null) {
bottomQueue.add(top.right);
}
}
result.add(level);
topQueue = bottomQueue;
}
return result;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-06-25 11:27
*/
public List<List<Integer>> levelOrder(TreeNode root) {
if (root == null) {
return Collections.emptyList();
}
List<List<Integer>> result = new ArrayList<>();
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
List<Integer> nums = new ArrayList<>(size);
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
nums.add(node.val);
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
result.add(nums);
}
return result;
}
| TODO：还可以优化成只用一个队列对象，思考如何实现？ |