{leetcode}/problems/populating-next-right-pointers-in-each-node/[LeetCode - 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:
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You may only use constant extra space.
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Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.
Example 1:
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:
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The number of nodes in the given tree is less than
4096. -
-1000 ⇐ node.val ⇐ 1000
这道题的关键是在上层遍历中，把下层的链接关系建立起来。
根据这个提示，再重新实现这个题目时，在思考的过程发现，最重要的就是保存每一层的最左节点。（这点也是根据树的 Morris 遍历得到的启发）这是每一层的起始位置。另外，每层循环的结束是当前节点为 null 了。（每层最后的一个节点没有 next 节点）。
因为是完全二叉树。所以，如果左下节点为空则到达最后一层；向右节点为空，则到达行尾需要换行。
link:{sourcedir}/_0116_PopulatingNextRightPointersInEachNode.java[role=include]