LeetCode117—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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本文详细介绍了如何使用层序遍历解决LeetCode117题，即在给定的任意二叉树中填充每个节点的next指针，使其相邻节点相连。通过使用常数额外空间，实现高效的解决方案。

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LeetCode117—Populating Next Right Pointers in Each Node II

原题

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
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分析

参考LeetCode116—Populating Next Right Pointers in Each Node直接bfs层序遍历即可

代码

class Solution {
private:
    void bfs(TreeLinkNode *root)
    {
        if (root == NULL)
            return;
        vector<TreeLinkNode*>temp;
        vector<TreeLinkNode*>q;
        int front = 0;
        int rear = 1;
        q.push_back(root);
        while (front < q.size())
        {
            rear = q.size();
            while (front < rear)
            {
                temp.push_back(q[front]);
                if (q[front]->left != NULL)
                    q.push_back(q[front]->left);
                if (q[front]->right != NULL)
                    q.push_back(q[front]->right);
                ++front;
            }
            int i;
            for ( i = 0; i < temp.size()-1; i++)
            {
                temp[i]->next = temp[i + 1];
            }
            temp[i]->next = NULL;
            temp.clear();//结束一层
        }
    }
public:
    void connect(TreeLinkNode *root) {
        bfs(root);
    }
};