[LeetCode] [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

/**
 * 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:
    TreeLinkNode *NextNode(TreeLinkNode *p) {
        while (p) {
            if (p->left) 
                return p->left;
            if (p->right)
                return p->right;
            p = p->next;
        }
        
        return NULL;
    }
    
    void connect(TreeLinkNode *root) {
        if (root == NULL) return;
        TreeLinkNode *level_begin = root;
        while (level_begin) {
            TreeLinkNode *cur = level_begin;
            while (cur) {
                if (cur->left)
                    cur->left->next = (cur->right != NULL) ? cur->right : NextNode(cur->next);
                if (cur->right)
                    cur->right->next = NextNode(cur->next);
                cur = cur->next;
            }
            level_begin = NextNode(level_begin); //下一层的开始节点
        }
    }
};

思路二：

　　一个prev记录当前层前一节点是啥（用来连接的
　　一个next记录下一层的开始（用户切换到下一层）

/**
 * 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) {
        while(root) {
            TreeLinkNode *next = NULL; //the first node of next level
            TreeLinkNode *prev = NULL; //previous node on the same level
            for (; root; root = root->next) {
                if (!next) 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;
        }
    }
};