方法1:BFS
时间复杂度:$O(n)$
空间复杂度:$O(n)$
解题思路
根节点入队
循环步骤1-3直到队列为空
1. 节点出队
2. 节点的左子节点入队
3. 节点的右子节点入队
Java 代码
/**
* 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>> res = new ArrayList<>();
if (root == null) {
return res;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
List<Integer> list = new ArrayList<>();
int levCnt = queue.size();
for (int i = 0; i < levCnt; i ++) {
TreeNode node = queue.remove();
list.add(node.val);
if (node.left != null) {
queue.add(node.left);
}
if (node.right != null) {
queue.add(node.right);
}
}
res.add(list);
}
return res;
}
}