这里的思路是N叉树的层序遍历
(递归法)
/*
// Definition for a Node.
class Node {
public int val;
public List<Node> children;
public Node() {}
public Node(int _val,List<Node> _children) {
val = _val;
children = _children;
}
};
public class levelOrder {
public static void main(String[] args) {
}
static List<List<Integer>> returnList = new ArrayList<>();
public static List<List<Integer>> levelOrder(Node root) {
if(root == null){
return returnList;
}
if(root.children == null){
returnList.get(0).add(root.val);
return returnList;
}
helper(root,0);
return returnList;
}
private static void helper(Node root,int index) {
if (returnList.size() == index) {
returnList.add(new ArrayList<>());
}
returnList.get(index).add(root.val);
for (int i = 0; i <root.children.size() ; i++) {
helper(root.children.get(i),index +1);
}
}
}
*/
队列(广度优先遍历)
public static void main(String[] args) {
Node node2 = new Node();
node2.val = 2;
Node node4 = new Node();
node4.val = 4;
Node node5 = new Node();
node5.val = 5;
Node node6 = new Node();
node6.val = 6;
List<Node> Node2 = new ArrayList<>();
Node2.add(node5);
Node2.add(node6);
Node node3 = new Node(3,Node2);
List<Node> Node1 = new ArrayList<>();
Node1.add(node3);
Node1.add(node2);
Node1.add(node4);
Node node1 = new Node(1,Node1 );
System.out.println(levelOrder(node1));
// List<List<Integer>> returnList = levelOrder(node1);
// for (int i = 0; i <returnList.size() ; i++) {
// for (int j = 0; j <returnList.get(i).size() ; j++) {
// System.out.print(returnList.get(i).get(j));
// }
// }
}
public static List<List<Integer>> levelOrder(Node root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) {
return res;
}
Queue<Node> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int count = queue.size();
//外层循环为一层
List<Integer> list = new ArrayList<>();
while (count-- > 0) {
//将当前元素的非空子节点压入栈
Node cur = queue.poll();
list.add(cur.val);
if(cur.children != null){
for (Node node : cur.children) {
if (node != null) {
queue.add(node);
}
}
}
}
res.add(list);
}
return res;
}
}
以上部分是N叉树的层序遍历
求最大的宽度
思路(其实就是先进行层次遍历,然后对所有的List进行遍历,计算最长的)
relationList为模拟的n叉树,appointSale为随机一个点。求这个随机点的子树下的最大宽度
private static int[] max(Map<Integer, List<Integer>> relationList, int appointSale) {
List<List<Integer>> res = new ArrayList<>();
LinkedList<Integer> ans = new LinkedList<>();
ans.add(appointSale);
while (!ans.isEmpty()){
int count = ans.size();
List<Integer> unFinds = new ArrayList<>();
while (count-- > 0){
int temp = ans.pop();
unFinds.add(temp);
if(relationList.containsKey(temp)){
List<Integer> tempList = relationList.get(temp);
for (int i = 0; i < tempList.size(); i++) {
ans.add(tempList.get(i));
}
}
}
res.add(unFinds);
}
int max = 0;
int index = 0;
for (int i = 0; i < res.size(); i++) {
if(res.get(i).size() > max){
max = Math.max(max,res.get(i).size());
index = i + 1;
}
}
int[] tempVec= new int[2];
tempVec[0] = index;
tempVec[1] = max;
return tempVec;
}
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