[Leetcode] Populating Next Right Pointers in Each Node II

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,

         1
       /  
      2    3
     /     
    4   5    7

After calling your function, the tree should look like:

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

跟上一题一样样的。

/**
 * 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 == NULL) {
            return;
        }
        queue<TreeLinkNode* > q;
        TreeLinkNode *p;
        int idx = 1, n;
        q.push(root);
        while (!q.empty()) {
            n = idx - 1;
            idx = 0;
            p = q.front();
           if (q.front()->left != NULL) {
                    q.push(q.front()->left);
                    idx++;
                }
                if (q.front()->right != NULL) {
                    q.push(q.front()->right);
                    idx++;
                }
            q.pop();
            for (int i = 0; i < n; ++i) {
                p->next = q.front();
                if (q.front()->left != NULL) {
                    q.push(q.front()->left);
                    idx++;
                }
                if (q.front()->right != NULL) {
                    q.push(q.front()->right);
                    idx++;
                }
                p = p->next;
                q.pop();
            }
            p->next = NULL;
        } 
    }
};

但是，如果只要常数的空间复杂度的话，我们可以利用next指针，因为在访问当前层时，当前行的next指针已经在访问上一层时连接好了。

/**
 * 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) {
        TreeLinkNode *prev, *next;
        while (root) {
            prev = nullptr; next = nullptr;
            for (; root != nullptr; root = root->next) {
                if (next == nullptr) next = root->left ? root->left : root->right;
                if (root->left) {
                    if (prev) prev->next = root->left;
                    prev = root->left;
                }
                if (root->right) {
                    if (prev) prev->next = root->right;
                    prev = root->right;
                }
            }
            root = next;
        }
    }
};