[LeetCode] 117. Populating Next Right Pointers in Each Node II
grandyang opened this issue · comments
Given a binary tree
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
.
Example:
Input: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":null,"next":null,"right":{"$id":"6","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
Output: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":null,"right":null,"val":7},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"6","left":null,"next":null,"right":{"$ref":"5"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"6"},"val":1}
Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.
Note:
- You may only use constant extra space.
- Recursive approach is fine, implicit stack space does not count as extra space for this problem.
这道是之前那道 Populating Next Right Pointers in Each Node 的延续,原本的完全二叉树的条件不再满足,但是整体的思路还是很相似,仍然有递归和非递归的解法。我们先来看递归的解法,这里由于子树有可能残缺,故需要平行扫描父节点同层的节点,找到他们的左右子节点。代码如下:
解法一:
class Solution {
public:
Node* connect(Node* root) {
if (!root) return NULL;
Node *p = root->next;
while (p) {
if (p->left) {
p = p->left;
break;
}
if (p->right) {
p = p->right;
break;
}
p = p->next;
}
if (root->right) root->right->next = p;
if (root->left) root->left->next = root->right ? root->right : p;
connect(root->right);
connect(root->left);
return root;
}
};
对于非递归的方法,我惊喜的发现之前的方法直接就能用,完全不需要做任何修改,算法思路可参见之前的博客 Populating Next Right Pointers in Each Node,代码如下:
解法二:
// Non-recursion, more than constant space
class Solution {
public:
Node* connect(Node* root) {
if (!root) return NULL;
queue<Node*> q;
q.push(root);
while (!q.empty()) {
int len = q.size();
for (int i = 0; i < len; ++i) {
Node *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);
}
}
return root;
}
};
虽然以上的两种方法都能通过OJ,但其实它们都不符合题目的要求,题目说只能使用constant space,可是OJ却没有写专门检测space使用情况的test,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cur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cur的next连上左子结点,然后cur指向其next指针;如果root的右子结点存在,那么cur的next连上右子结点,然后cur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cur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,或者也可以将cur的next指针清空,因为前面已经将cur赋值为dummy了。那么现在想一想,为什么要清空?因为我们用dummy的目的就是要指到下一行的首结点的位置即dummy->next,而一旦将root赋值为dummy->next了之后,这个dummy的使命就已经完成了,必须要断开,如果不断开的话,那么假设现在root是叶结点了,那么while循环还会执行,不会进入前两个if,然后root右移赋空之后,会进入最后一个if,之前没有断开dummy->next的话,那么root又指向之前的叶结点了,死循环诞生了,跪了。所以一定要记得清空哦,呵呵哒~
这里再来说下dummy结点是怎样指向每层的首结点的前一个结点的,过程是这样的,dummy是创建出来的一个新的结点,其目的是为了指向root结点的下一层的首结点的前一个,具体是这么做到的呢,主要是靠cur指针,首先cur指向dummy,然后cur再连上root下一层的首结点,这样dummy也就连上了。然后当root层遍历完了之后,root需要往下移动一层,这样dummy结点之后连接的位置就正好赋值给root,然后cur再指向dummy,dummy之后断开,这样又回到了初始状态,以此往复就可以都连上了,代码如下:
解法三:
// Non-recursion, constant space
class Solution {
public:
Node* connect(Node* root) {
Node *dummy = new Node(0, NULL, NULL, NULL), *cur = dummy, *head = root;
while (root) {
if (root->left) {
cur->next = root->left;
cur = cur->next;
}
if (root->right) {
cur->next = root->right;
cur = cur->next;
}
root = root->next;
if (!root) {
cur = dummy;
root = dummy->next;
dummy->next = NULL;
}
}
return head;
}
};
类似题目:
Populating Next Right Pointers in Each Node
参考资料:
https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/