给定一个 N 叉树,返回其节点值的层序遍历。(即从左到右,逐层遍历)。
树的序列化输入是用层序遍历,每组子节点都由 null 值分隔(参见示例)。
示例 1:
输入:root = [1,null,3,2,4,null,5,6]
输出:[[1],[3,2,4],[5,6]]
示例 2:
输入:root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
输出:[[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]
提示:
- 树的高度不会超过
1000
- 树的节点总数在
[0, 10^4]
之间
题解:
直接进行遍历,用临时变量保存所有下一层节点。
code:
java
/*
// Definition for a Node.
class Node {
public int val;
public List<Node> children;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, List<Node> _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public List<List<Integer>> levelOrder(Node root) {
List<List<Integer>> result = new ArrayList<>();
if (root == null) {
return result;
}
boolean isContinue = true;
List<Node> tmp = new ArrayList<Node>();
List<Node> tmp2 = new ArrayList<Node>();
tmp.add(root);
while(isContinue) {
List<Integer> list = new ArrayList<Integer>();
for(Node node : tmp) {
if (node == null) {
continue;
}
list.add(node.val);
tmp2.addAll(node.children);
}
result.add(list);
if (tmp2.isEmpty()) {
isContinue = false;
}
tmp = tmp2;
tmp2 = new ArrayList<Node>();
}
return result;
}
}
使用队列实现
java
public List<List<Integer>> levelOrder(Node root) {
if (root == null) {
return new ArrayList<List<Integer>>();
}
List<List<Integer>> ans = new ArrayList<List<Integer>>();
Queue<Node> queue = new ArrayDeque<Node>();
queue.offer(root);
while (!queue.isEmpty()) {
int cnt = queue.size();
List<Integer> level = new ArrayList<Integer>();
for (int i = 0; i < cnt; ++i) {
Node cur = queue.poll();
level.add(cur.val);
for (Node child : cur.children) {
queue.offer(child);
}
}
ans.add(level);
}
return ans;
}