LeetCode | Populating Next Right Pointers in Each Node II（将每一层链接成一个链表）

2023年2月15日 240次阅读

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,

         1
       /  \
      2    3
     / \    \
    4   5    7

After calling your function, the tree should look like:

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

题目解析：

本来这道题有个完全二叉树的，但感觉没必要，利用两种通用的方法都能解决，这里只写这道题了。

方案一：

利用程序遍历的思想，每一层之间的元素链接成链表，循环到队列为空为止。

class Solution {
public:
    void connect(TreeLinkNode *root) {
        if(root == NULL)
            return ;
        queue<TreeLinkNode *> qtree;
        qtree.push(root);
        while(!qtree.empty()){
            int n = qtree.size();
            TreeLinkNode *p = qtree.front();
            qtree.pop();    //记得要出队列，同时把子孩子入队列
            if(p->left) qtree.push(p->left);
            if(p->right) qtree.push(p->right);
            for(int i = 1;i < n;i++){
                TreeLinkNode *q = qtree.front();
                qtree.pop();
                if(q->left) qtree.push(q->left);
                if(q->right) qtree.push(q->right);
                p->next = q;
                p = q;
            }
            p->next = NULL;
        }
    }
};

方案二：

很巧妙的方法，其实也是一层一层的去链接，但是当处理第i层的时候，i-1层已经链接成一个链表，直接用cur = cur->next，就能找到下一对结点的位置。充分利用了树的结构。

/**
 * 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 ;
        root->next = NULL;
        TreeLinkNode* prevHead = root;
        while(true)
        {
            TreeLinkNode* cur = prevHead;
            TreeLinkNode* prev = NULL;
            while(cur != NULL)
            {
                if(cur->left != NULL)
                {
                    if(prev != NULL) prev->next = cur->left, prev = prev->next;
                    else prev = cur->left, prevHead = prev;
                }
                if(cur->right != NULL)
                {
                    if(prev != NULL) prev->next = cur->right, prev = prev->next;
                    else prev = cur->right, prevHead = prev;
                }
                cur = cur->next;
            }
            if(prev != NULL) prev->next = NULL;
            else break;
        }
    }
};


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