102. Binary Tree Level Order Traversal
102. Binary Tree Level Order Traversal - Medium
Description
102. Binary Tree Level Order Traversal
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]]
这道题很经典的一道二叉树的层序遍历，套用BFS模版就行。
Solution
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35 | /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
if (root == null) return new ArrayList<>();
Deque<TreeNode> q = new LinkedList<>();
q.offer(root);
List<List<Integer>> ans = new ArrayList<>();
while (!q.isEmpty()) {
int size = q.size();
List<Integer> level = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode node = q.poll();
level.add(node.val);
if (node.left != null) q.offer(node.left);
if (node.right != null) q.offer(node.right);
}
ans.add(level);
}
return ans;
}
}
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