Populating Next Right Pointers in Each Node II

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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CC 4.0 BY-SA版权

文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/memoryh/p/3841206.html

题目地址： https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

关键思路：讲节点的左右子节点链接起来，遍历节点的next节点，即可。

class Solution {
public:
    TreeLinkNode* gethead(TreeLinkNode* root)
    {
        if(root == NULL)
            return NULL;
        if(root->left != NULL)
            return root->left;
        else
            return root->right;
    }
    
    TreeLinkNode* gettail(TreeLinkNode* root)
    {
        if(root == NULL)
            return NULL;
        if(root->right != NULL)
            return root->right;
        else
            return root->left;
    }

    void connect(TreeLinkNode *root) 
    {    
        TreeLinkNode* head = NULL;
        TreeLinkNode* tail = NULL;
        if(root == NULL)
            return;

        TreeLinkNode* pcur = root;
        while(root != NULL)
        {
            TreeLinkNode* headtmp = gethead(root);
            TreeLinkNode* tailtmp = gettail(root);

            if(headtmp == NULL)
            {
                root= root->next;
                continue;
            }

            if(headtmp != tailtmp)
                headtmp->next = tailtmp;

            if(tail != NULL)
            {
                tail->next = headtmp;
                tail = tailtmp;
            }
            else
            {
                head = headtmp;
                tail = tailtmp;
            }
            root = root->next;
        }

        while(pcur != NULL)
        {
            TreeLinkNode* tmp = gethead(pcur);
            if(tmp != NULL)
            {
                connect(tmp);
                break;
            }
            pcur = pcur->next;
        }
    }

};