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.
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.
1 2 3
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.
The number of nodes in the given tree is less than 4096.
-1000 <= node.val <= 1000
对于数组中的节点 n (下标为 i), 如果存在 n1 (下标为 i + 1), 则 n1 一定是 n 的右侧节点.