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429. N-ary Tree Level Order Traversal
Given an n-ary tree, return the level order traversal of its nodes" values. (ie, from left to right, level by level).
For example, given a 3-ary tree:
We should return its level order traversal:
[
[1],
[3,2,4],
[5,6]
]
Note:
The depth of the tree is at most 1000.
The total number of nodes is at most 5000.
use Queue, in each level, use queue.size() to loop
class Solution {
public List> levelOrder(Node root) {
List> res = new ArrayList<>();
if (root == null) return res;
Queue queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
List curRes = new ArrayList<>();
int size = queue.size();
for (int i = 0; i < size; i++) {
Node node = queue.poll();
curRes.add(node.val);
for (Node child: node.children) {
queue.offer(child);
}
}
res.add(curRes);
}
return res;
}
}
102. 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] ]Solution (Iteration)
class Solution {
public List> levelOrder(TreeNode root) {
List> res = new ArrayList<>();
if (root == null) return res;
Queue queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
List curRes = new ArrayList<>();
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
curRes.add(node.val);
if (node.left != null) queue.offer(node.left);
if (node.right != null) queue.offer(node.right);
}
res.add(curRes);
}
return res;
}
}
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