LeetCode 117: Populating Next Right Pointers in Each Node II

最新推荐文章于 2022-06-02 20:32:40 发布

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本文介绍了如何使用队列或仅使用next指针来解决任意二叉树层次遍历问题，提供了两种不同的实现方法，旨在为解决实际问题提供灵活的策略。

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,

After calling your function, the tree should look like:

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

解题思路

与 LeetCode 116 不同的是本题的输入可以是任意二叉树。

思路一：依然可以利用一个队列（Queue）对二叉树进行层次遍历，在遍历过程中将同一层的节点通过 next 之间链接起来。所需空间不是 constant extra space。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) {
         if (root == NULL) return;

         queue<TreeLinkNode*> nodeQueue;
         nodeQueue.push(root);
         while (!nodeQueue.empty()) {
             int levelSize = nodeQueue.size();

             for (int i = 0; i < levelSize; ++i) {
                 TreeLinkNode *p = nodeQueue.front();
                 nodeQueue.pop();

                 if (i < levelSize - 1) {
                     p->next = nodeQueue.front();
                 }

                 if (p->left != NULL) nodeQueue.push(p->left);
                 if (p->right != NULL) nodeQueue.push(p->right);
             }
         }
    }
};

思路二：借用 next 指针，做到不需要队列就能完成广度优先搜索（BFS）：如果当前层所有结点的 next 指针已经设置好了，那么据此，下一层所有结点的 next 指针也可以依次被设置。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) {
         while (root != NULL) {
             TreeLinkNode *p = root, *first = NULL, *pre = NULL;
             while (p != NULL) {
                 if (p->left != NULL) {
                     if (first == NULL) 
                        first = p->left;

                     if (pre != NULL) {
                         pre->next = p->left;
                     }
                     pre = p->left;
                 }

                 if (p->right != NULL) {
                     if (first == NULL) 
                        first = p->right;

                     if (pre != NULL) {
                         pre->next = p->right;
                     }
                     pre = p->right;
                 }

                 p = p->next;
             }
             root = first;
         }
    }
};