LeetCode: Populating Next Right Pointers in Each Node II

这题关键是要记录下一层第一个节点

/**
 * 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:
    TreeLinkNode *nextright(TreeLinkNode *root) {
        while (root) {
            if (root->left) return root->left;
            if (root->right) return root->right;
            root = root->next;
        }
        return NULL;
    }
    void connect(TreeLinkNode *root) {
        if (!root) return;
        TreeLinkNode *second, *first;
        first = root;
        second = NULL;
        first->next = NULL;
        while (first) {
            if (first->left) {
                if (!second) second = first->left;
                if (first->right) first->left->next = first->right;
                else first->left->next = nextright(first->next);
            }
            if (first->right) {
                if (!second) second = first->right;
                first->right->next = nextright(first->next);
            }
            first = first->next;
            if (!first) {
                first = second;
                second = NULL;
            }
        }
    }
};

 如果可以用extra space，可以用下面这段代码

/**
 * 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) {
        queue<TreeLinkNode *> S;
        if (!root) return;
        S.push(root);
        S.push(NULL);
        while (!S.empty()) {
            TreeLinkNode *front = S.front();
            S.pop();
            if (!front) continue;
            front->next = S.front();
            if (front->left) S.push(front->left);
            if (front->right) S.push(front->right);
            if (!S.front()) S.push(NULL);
        }
    }
};