Problem description:

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:

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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

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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.

Solution BFS:

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"""
# Definition for a Node.
class Node:
    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""

class Solution:
    def connect(self, root: 'Node') -> 'Node':
        if not root:
            return root
        res, queue = [], deque([root])
        while queue:
            size = len(queue)
            next_level = deque([])
            for _ in range(size):
                node = queue.popleft()
                if queue: 
                    node.next = queue[0]
                if node.left: next_level.append(node.left)
                if node.right: next_level.append(node.right)
            queue = next_level
        return root
```        
        
## Solution DFS:
We have `root.next` setup for root node, so the `root.right.next` would be `root.next.left`.

```python
class Solution:
    def connect(self, root: 'Node') -> 'Node':
        def dfs(root):
            if not root:
                return root
            if root.right:
                root.left.next = root.right
                if root.next:
                    root.right.next = root.next.left
            dfs(root.left)
            dfs(root.right)
        dfs(root)
        return root

time complexity: $O()$
space complexity: $O()$
reference:
related problem: