[leetcode-117]Populating Next Right Pointers in Each Node II(c++)

问题描述：
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：代码与上一道题基本完全一样。

代码如下：40ms

/** * 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)            return;        queue<TreeLinkNode *> queue;        queue.push(root);        int count = 1;        while (!queue.empty()) {            int tmpCount = 0;            TreeLinkNode* head = NULL;            for (int i = 0; i < count; i++) {                TreeLinkNode* tmp = queue.front();                queue.pop();                if(tmp->left){                    tmpCount++;                    queue.push(tmp->left);                }                if(tmp->right){                    tmpCount++;                    queue.push(tmp->right);                }                if (!head)                    head = tmp;                else {                    head->next = tmp;                    head = tmp;                }            }            count = tmpCount;        }    }};