[leetcode 117] Populating Next Right Pointers in Each Node II

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本文讨论了如何在给定的复杂二叉树结构中实现节点链接，包括节点的左右子节点以及兄弟节点之间的链接。通过使用递归算法，实现了一种仅使用常量额外空间的方法来构建链接，使得树的结构可以方便地进行遍历和操作。

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.

For example,
Given the following binary tree,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL

递归

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        if (!root) {
            return ;
        }
        TreeLinkNode dummy(-1);
        for (TreeLinkNode *cur = root, *prev = &dummy; cur; cur = cur->next) {
            if (cur->left) {
                prev->next = cur->left;
                prev = cur->left;
            }
            if (cur->right) {
                prev->next = cur->right;
                prev = cur->right;
            }
        }
        connect(dummy.next);
    }
};