【To Understand!】LeetCode 117. Populating Next Right Pointers in Each Node II

LeetCode 117. Populating Next Right Pointers in Each Node II

Solution1：我的答案
层次遍历

/**
 * 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;
        queue<TreeLinkNode*> q;
        q.push(root);
        int cur = 1, next = 0;
        while (!q.empty()) {
            TreeLinkNode* temp = q.front();
            q.pop();
            if (cur > 1)
                temp->next = q.front();
            cur--;
            if (temp->left) {
                q.push(temp->left);
                next++;
            } 
            if (temp->right) {
                q.push(temp->right);
                next++;
            } 
            if (cur == 0) {
                cur = next;
                next = 0;
            }
        }
    }
};

Solution2:
参考：转载自：http://www.cnblogs.com/grandyang/p/4290148.html
层次遍历，简练的写法值得学习一个！

/**
 * 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;
        queue<TreeLinkNode*> q;
        q.push(root);
        while (!q.empty()) {
            int len = q.size();
            for (int i = 0; i < len; ++i) {
                TreeLinkNode *t = q.front(); q.pop();
                if (i < len - 1) t->next = q.front();
                if (t->left) q.push(t->left);
                if (t->right) q.push(t->right);
            }
        }
    }
};

【有待理解啊】Solution3:
参考：转载自：http://www.cnblogs.com/grandyang/p/4290148.html
那么下面贴上constant space的解法，这个解法也是用的层序遍历，只不过没有使用queue了，我们建立一个dummy结点来指向每层的首结点的前一个结点，然后指针t用来遍历这一层，我们实际上是遍历一层，然后连下一层的next，首先从根结点开始，如果左子结点存在，那么t的next连上左子结点，然后t指向其next指针；如果root的右子结点存在，那么t的next连上右子结点，然后t指向其next指针。此时root的左右子结点都连上了，此时root向右平移一位，指向其next指针，如果此时root不存在了，说明当前层已经遍历完了，我们重置t为dummy结点，root此时为dummy->next，即下一层的首结点，然后dummy的next指针清空，代码如下：

/**
 * 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 *dummy = new TreeLinkNode(0), *t = dummy;
        while (root) {
            if (root->left) {
                t->next = root->left;
                t = t->next;
            }
            if (root->right) {
                t->next = root->right;
                t = t->next;
            }
            root = root->next;
            if (!root) {
                t = dummy;
                root = dummy->next;//dummy存为t的起点，当t指向下一层的首结点时，dummy也指了过去？？
                dummy->next = NULL;
            }
        }
        return;
    }
};