# 116. Populating Next Right Pointers in Each Node

[Link](https://leetcode.com/problems/populating-next-right-pointers-in-each-node/)

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`.

**Follow up:**

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

**Example 1:**

![](https://assets.leetcode.com/uploads/2019/02/14/116_sample.png)

```
Input: root = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]
Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.
```

**Constraints:**

* The number of nodes in the given tree is less than `4096`.
* `-1000 <= node.val <= 1000`

**Solution**

**Recursion**\
把每個子樹的connection建立起來 connect建立的順序要注意，先建立內部左右節點的

```
connectNode(node2.left, node2.right);
            connectNode(node1.left, node1.right);
```

再建立 兩子樹sibling之間的

```
connectNode(node1.right, node2.left);
```

T:O(N)\
S: O(logN), 最高call stack 數

```
 public Node connect(Node root) {
        if(root == null) return root;
        root.next = null;
        if(root.left != null){
            connectNode(root.left, root.right);
        }
        return root;
    }
    public void connectNode(Node node1, Node node2) {
        node1.next = node2;
        if(node1.left != null){
            connectNode(node2.left, node2.right);
            connectNode(node1.left, node1.right);
            connectNode(node1.right, node2.left);
        }
    }
```

**Level traversals**

因為有next pointer的幫助，變成可以不用queue去記載。

T: O(N)\
S: O(1)

```
class Solution {
    public Node connect(Node root) {
        Node start = root;
        while(start != null){
            //do it by level
            Node currentNode = start;
            while(currentNode != null){
                if(currentNode.left != null){
                    currentNode.left.next = currentNode.right;
                    if(currentNode.next != null){
                        currentNode.right.next = currentNode.next.left;
                    }
                }
                currentNode = currentNode.next;
            }
            start = start.left;//to Next level
        }
        return root;
    }
}
```
