[Leetcode] 117. Populating Next Right Pointers in Each Node II 解题报告_117 populating next right pointers in each node ||-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/70882654

本文探讨了如何在常数额外空间限制下解决二叉树节点填充问题，介绍了一种巧妙的方法来连接同一层的节点，使其指向右侧相邻的节点。通过使用虚拟头结点跟踪每层的第一个节点，并迭代地连接所有兄弟节点。

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题目：

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

思路：

我原来的做法是采用队列进行层次遍历，对同一层次的元素依次添加即可。然而这种做法的空间复杂度是O(n)量级的，所以不符合题目要求。

在网上参考了别人的代码，觉得很巧妙：利用一个虚拟头结点，记录要连接指针的一层节点的第一个元素，然后一个指针从左到右依次连接上一层节点的左右节点。

代码：

/**
 * 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 *pHead = new TreeLinkNode(0), *pre = pHead;  
        while (root) {  
            if(root->left)  
            {  
                pre->next = root->left;  
                pre = pre->next;  
            }  
            if(root->right)  
            {  
                pre->next = root->right;  
                pre = pre->next;  
            }  
            root = root->next;  
            if (!root)          // come to the end of one layer
            {  
                pre = pHead;  
                root = pHead->next;  
                pHead->next = NULL;  
            }  
        }
        delete pHead;
    }
};