117. Populating Next Right Pointers in Each Node II

O(n) time O(1) space

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/

class Solution {
public:
    Node* connect(Node* root) {
        Node dummy, *p = &dummy, *res = root;
        while (root) {
            if (root->left) {
                p->next = root->left;
                p = p->next;
            }
            if (root->right) {
                p->next = root->right;
                p = p->next;
            }
            root = root->next; // 假设上一层已经连完了
            if (!root) { // 到头了
                root = dummy.next; // 设成下一层起始
                dummy.next = nullptr; // 重置dummy和p
                p = &dummy;
            }
        }
        return res;
    }
};

dfs O(n) time O(h) space
从上往下连，从右往左连，因为上边的连好了下边的才能利用上边的连，右边的连好了左边的才不会连错

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() {}

    Node(int _val, Node* _left, Node* _right, Node* _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/
class Solution {
public:
    Node* connect(Node* root) {
        if (!root) return root;
        Node dummy_head(0, nullptr, nullptr, nullptr), *tail = &dummy_head;
        for (auto p = root; p; p = p->next) {
            if (p->left) {
                tail->next = p->left;
                tail = tail->next;
            }
            if (tail->next) break; // 已经连上的没必要再连一遍
            if (p->right) {
                tail->next = p->right;
                tail = tail->next;
            }
            if (tail->next) break;
        }
        connect(root->right);
        connect(root->left);
        return root;
    }
};

/**
 * 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;
        TreeLinkNode dummy_head(0), *curr = &dummy_head; // 利用dummy_head实现『盲连』，即避免不必要的节点判断
        for (auto p = root; p; p = p->next) { // suppose当前层已经连好了，只需要把子一层的都连起来
            if (p->left) {
                curr->next = p->left;
                curr = p->left;
            }
            if (p->right) {
                curr->next = p->right;
                curr = p->right;
            }
        }
        connect(root->right);
        connect(root->left);
    }
};

/**
 * 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;
        if (root->left) {
            if (root->right) {
                root->left->next = root->right;
            } else {
                root->left->next = next(root->next);
            }
        }
        if (root->right) {
            root->right->next = next(root->next);
        }
        connect(root->right); // 一定要先右后左！！！！！先左后右有一些会没连上
        connect(root->left);
    }

    TreeLinkNode *next(TreeLinkNode *root) {
        while (root) {
            if (root->left) return root->left;
            if (root->right) return root->right;
            root = root->next;
        }
        return nullptr;
    }
};