Tree——No.117 Populating Next Right Pointers in Each Node II

Problem：

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

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.

Explanation：

将每个结点的next指向它右边的结点，和116不同的是116是完全二叉树，而本题存在null。

My Thinking：

和116一样进行层序遍历，然后改变指针，由于116次层序遍历已经去除了结点为null的情况，因此直接使用即可。

My Solution：

class Solution {
    public Node connect(Node root) {
        if(root==null)
            return root;
        Queue<Node> queue=new LinkedList<>();
        queue.add(root);
        
        while(!queue.isEmpty()){
            int length=queue.size();
            Node node1=new Node();
            for(int i=0;i<length;i++){
                Node node2=queue.poll();
                if(node2.left!=null)
                    queue.add(node2.left);
                if(node2.right!=null)
                    queue.add(node2.right);
                node1.next=node2;
                node1=node2;
            }
            node1.next=null;
        }
        return root;
    }
}

Optimum Thinking:

Optimum Solution：