LeetCode: Populating Next Right Pointers in Each Node

思路对的，少数次过，基本一次过吧

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        queue<TreeLinkNode *> S;
        queue<TreeLinkNode *> T;
        if (!root) return;
        S.push(root);
        while (!S.empty()) {
            TreeLinkNode *tmp = S.front();
            S.pop();
            if (tmp->left || tmp->right) {
                T.push(tmp->left);
                T.push(tmp->right);
            }
            if (!S.empty()) tmp->next = S.front();
            else tmp->next = NULL;
            if (S.empty()) {
                while (!T.empty()) {
                    S.push(T.front());
                    T.pop();
                }
            }
        }
    }
};

 贴一段后来写的更加好的代码

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (!root) return;
        queue<TreeLinkNode*> S, T;
        S.push(root);
        while (!S.empty()) {
            while (!S.empty()) {
                TreeLinkNode *tmp = S.front();
                S.pop();
                if (tmp->left) T.push(tmp->left);
                if (tmp->right) T.push(tmp->right);
                tmp->next = S.empty()? NULL : S.front();
            }
            S.swap(T);
        }
    }
};