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`

.

**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":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
Output: {"$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}
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.
```

**Note**:

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

## Solutions

### 1. DFS-先序遍历-递归

采用递归的方法，注意这里的递归条件。

```
# Time Complexity: O(n)
# Space Complexity: O(logn)
"""
# Definition for a Node.
class Node:
def __init__(self, val, left, right, next):
self.val = val
self.left = left
self.right = right
self.next = next
"""
class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return None
if not root.left:
return root
root.left.next = root.right
if root.next:
root.right.next = root.next.left
root.left = self.connect(root.left)
root.right = self.connect(root.right)
return root
# Runtime: 60 ms, faster than 98.51% of Python3 online submissions for Populating Next Right Pointers in Each Node.
# Memory Usage: 14.8 MB, less than 100.00% of Python3 online submissions for Populating Next Right Pointers in Each Node.
```

### 2. BFS-迭代

```
# Time Complexity: O(n)
# Space Complexity: O(logn)
"""
# Definition for a Node.
class Node:
def __init__(self, val, left, right, next):
self.val = val
self.left = left
self.right = right
self.next = next
"""
class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return None
queue = collections.deque([root])
while queue:
size = len(queue)
pre = None
for i in range(size):
node = queue.popleft()
if not pre:
pre = node
else:
pre.next = node
pre = node
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return root
# Runtime: 64 ms, faster than 93.22% of Python3 online submissions for Populating Next Right Pointers in Each Node.
# Memory Usage: 14.8 MB, less than 100.00% of Python3 online submissions for Populating Next Right Pointers in Each Node.
```