116. Populating Next Right Pointers in Each Node

116. Populating Next Right Pointers in Each Node

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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

1. 使用BFS遍历整棵树。
2. 这个方法的关键点在于，将每一层结尾的NULLpush入队列(以读到上一层结尾的NULL为标志 push)。
3. 如果出现连续的NULL说明遍历结束。

算法复杂度

$O(|V+E|)$

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
      }
    }
  }
};