102. Binary Tree Level Order Traversal

Question

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).

Solution

BFS搜索,层序遍历,记录每个层级的节点数量size。
在遍历时挤出size个节点,并将其子节点加入队列。

Code

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/**
* 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) {
List<List<Integer>> ret = new ArrayList<>();
if(root == null) return ret;
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);

int size = 1;
while(!q.isEmpty()){
List<Integer> list = new ArrayList<>();
TreeNode curr = null;
for(int i = 0; i < size; i++){
curr = q.poll();
if(curr.left != null) q.offer(curr.left);
if(curr.right != null) q.offer(curr.right);
list.add(curr.val);
}
ret.add(list);
size = q.size();
}
return ret;
}
}
Author

Xander

Posted on

2022-04-13

Updated on

2022-07-12

Licensed under

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