leetcode：116. 填充每个节点的下一个右侧节点指针（java树层次遍历bfs）

傷城~ 2021-09-28 12:38 161阅读 0赞

package LeetCode;
    
    import java.util.LinkedList;
    import java.util.Queue;
    /*
    给定一个完美二叉树，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：
    struct Node {
      int val;
      Node *left;
      Node *right;
      Node *next;
    }
    填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。
    初始状态下，所有 next 指针都被设置为 NULL。
    示例：
    输入：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
    输出：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}
    解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。
     */
    class Node {
        public int val;
        public Node left;
        public Node right;
        public Node next;
    
        public Node() {
        }
    
        public Node(int _val, Node _left, Node _right, Node _next) {
            val = _val;
            left = _left;
            right = _right;
            next = _next;
        }
    }
    
    public class Connect {
        /*
        思路：用层次遍历然后先把第一个取出然后遍历剩余的节点时用next连起来即可
         */
        public Node connect(Node root) {
            if(root==null) return null;
            Queue<Node> queue = new LinkedList<>();
            queue.add(root);
            while (!queue.isEmpty()) {
                int qcount = queue.size();
                qcount--;
                Node first = queue.poll();
                if (first.left != null) {
                    queue.add(first.left);
                }
                if (first.right != null) {
                    queue.add(first.right);
                }
                while (qcount > 0) {
                    Node now = queue.poll();
                    if (now.left != null) {
                        queue.add(now.left);
                    }
                    if (now.right != null) {
                        queue.add(now.right);
                    }
                    first.next = now;
                    first = first.next;
                    qcount--;
                }
            }
            return root;
        }
    
        public static void main(String[] args) {
            Node node4 = new Node(4, null, null, null);
            Node node5 = new Node(5, null, null, null);
            Node node6 = new Node(6, null, null, null);
            Node node7 = new Node(7, null, null, null);
            Node node2 = new Node(2, node4, node5, null);
            Node node3 = new Node(3, node6, node7, null);
            Node node1 = new Node(1, node2, node3, null);
            Connect connect = new Connect();
            connect.connect(node1);
    
        }
    }