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],
return its level order traversal as: [ [3], [9,20], [15,7] ]
思路
用宽度优先搜索的方法, 用while循环遍历树的每一层, for循环做每一层内的遍历
复杂度
时间 O(n) 空间O(n)
代码
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) {
return res;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
List<Integer> level = new ArrayList<>();
for (int i =0; i< size; i++) {
TreeNode node = queue.poll();
level.add(node.val);
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
res.add(level);
}
return res;
}
}
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