116. 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:

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

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本题要给二叉树的每个节点添加next指针，指向它的右兄弟，如果没有右兄弟，则为NULL。 本题使用层次遍历可以很容易解决，但是需要使用额外的空间，不符合题目要求。 根据题目中的示意图，很明显如果节点是父亲的左孩子，则其next指向父亲的右孩子；但是如果节点是父亲的右孩子，则next指向需要跨界，不太好办，经题解提示，这种情况下，next指向父亲的next的左孩子，真是巧妙呀。所以可以用DFS完成本题，代码如下：

class Solution {
public:
    void dfs(TreeLinkNode* child, TreeLinkNode* parent)
    {
        if (child == NULL)
            return;
        if (parent) {
            if (child == parent->left)
                child->next = parent->right;
            else if (parent->next)
                child->next = parent->next->left;
        }
        dfs(child->left, child);
        dfs(child->right, child);
    }
    void connect(TreeLinkNode* root) { dfs(root, NULL); }
};

本代码提交AC，用时26MS。