116. Populating Next Right Pointers in Each Node

116. Populating Next Right Pointers in Each Node

Given a binary tree

struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *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

.

Note:

- You may only use constant extra space.

- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,

Given the following perfect binary tree,

1
/  \
2    3
/ \  / \
4  5  6  7

After calling your function, the tree should look like:

1 -> NULL
/  \
2 -> 3 -> NULL
/ \  / \
4->5->6->7 -> NULL

题目大意

给定一个二叉树是一个完美二叉树(每个父节点都有两个子节点)。

将其每层上的节点从左到右使用next指针连接起来，如果没有右边的节点，则用NULL代替。

解题思路

使用BFS遍历整棵树。

这个方法的关键点在于，将每一层结尾的

NULL

push入队列(以读到上一层结尾的

NULL

为标志 push)。

如果出现连续的

NULL

说明遍历结束。

算法复杂度

O(|V+
4000
E|)

代码实现

/**
* 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) {
queue<TreeLinkNode *> que;
TreeLinkNode *cache = nullptr;
TreeLinkNode *top = nullptr;
que.push(root);
que.push(nullptr);  // key
// bfs
while (!que.empty()) {
top = que.front();
que.pop();
if (top != nullptr) {
top->next = que.front();
que.push(top->left);
que.push(top->right);
} else {
// 如果出现连续的nullptr, 则说明bfs已经结束
if (que.front() == nullptr) break;
que.push(nullptr);  // 层结束标志
}
}
}
};