leetcode116. 填充每个节点的下一个右侧节点指针

116. 填充每个节点的下一个右侧节点指针

难度中等128

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。

提示：

• 你只能使用常量级额外空间。
• 使用递归解题也符合要求，本题中递归程序占用的栈空间不算做额外的空间复杂度。

思路：层序遍历稍微改一下，把相同层的连一下即可。

  

   

    

     

    
    

     

      /*
     
    
   

    

     

    
    

     

      // Definition for a Node.
     
    
   

    

     

    
    

     

      class Node {
     
    
   

    

     

    
    

     

       public int val;
     
    
   

    

     

    
    

     

       public Node left;
     
    
   

    

     

    
    

     

       public Node right;
     
    
   

    

     

    
    

     

       public Node next;
     
    
   

    

     

    
    

     

      
     
    
   

    

     

    
    

     

       public Node() {}
     
    
   

    

     

    
    

     

       
     
    
   

    

     

    
    

     

       public Node(int _val) {
     
    
   

    

     

    
    

     

       val = _val;
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     

      
     
    
   

    

     

    
    

     

       public Node(int _val, Node _left, Node _right, Node _next) {
     
    
   

    

     

    
    

     

       val = _val;
     
    
   

    

     

    
    

     

       left = _left;
     
    
   

    

     

    
    

     

       right = _right;
     
    
   

    

     

    
    

     

       next = _next;
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     

      };
     
    
   

    

     

    
    

     

      */
     
    
   

    

     

    
    

     

      class Solution {
     
    
   

    

     

    
    

     
 public Node connect(Node root) {
     
    
   

    

     

    
    

     
 if (root == null) {
     
    
   

    

     

    
    

     
 return null;
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     

       Queue<Node> Q = new LinkedList<Node>(); 
     
    
   

    

     

    
    

     

       Q.add(root);
     
    
   

    

     

    
    

     
 while (Q.size() > 0) {
     
    
   

    

     

    
    

     
 int size = Q.size();
     
    
   

    

     

    
    

     
 for(int i = 0; i < size; i++) {
     
    
   

    

     

    
    

     

       Node node = Q.poll();
     
    
   

    

     

    
    

     
 if (i < size - 1) {
     
    
   

    

     

    
    

     

       node.next = Q.peek();
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     
 if (node.left != null) {
     
    
   

    

     

    
    

     

       Q.add(node.left);
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     
 if (node.right != null) {
     
    
   

    

     

    
    

     

       Q.add(node.right);
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     
 return root;
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     

      }
     
    
  
 

文章来源: fantianzuo.blog.csdn.net，作者：兔老大RabbitMQ，版权归原作者所有，如需转载，请联系作者。

原文链接：fantianzuo.blog.csdn.net/article/details/104371746