117. Populating Next Right Pointers in Each Node II

Given a binary tree

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Example:

Input: {"$id":"1","left":{"$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":{"$id":"5","left":null,"next":null,"right":{"$id":"6","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

Output: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":null,"right":null,"val":7},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"6","left":null,"next":null,"right":{"$ref":"5"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"6"},"val":1}

Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.

Note:

• You may only use constant extra space.
• Recursive approach is fine, implicit stack space does not count as extra space for this problem.

  class Solution {
      public Node connect(Node root) {
          if(root == null) return null;
          Queue<Node> q = new LinkedList();
          q.offer(root);
          while(!q.isEmpty()){
              Node pre = null;
              int s = q.size();
              for(int i = 0; i < s; i++){
                  Node cur = q.poll();
                  if(i > 0){
                      pre.next = cur;
                  }
                  pre = cur;
                  if(cur.left != null) q.offer(cur.left);
                  if(cur.right != null) q.offer(cur.right);
              }
          }
          return root;
      }
  }

  利用一个栈将下一层的节点保存。通过pre指针把栈里的元素一个一个接起来。可用于116题