117. Populating Next Right Pointers in Each Node II

Given a binary tree

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Example:

Input: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":null,"next":null,"right":{"$id":"6","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

Output: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":null,"right":null,"val":7},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"6","left":null,"next":null,"right":{"$ref":"5"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"6"},"val":1}

Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.

Note:

• You may only use constant extra space.

• Recursive approach is fine, implicit stack space does not count as extra space for this problem.

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;
    }
};

// O(n) Extra Space
Node* connect(Node* root) { // time: O(n); space: O(n)
    if (!root) return root;
    queue<Node*> q({root});
    while (!q.empty()) {
        for (int i = q.size() - 1; i >= 0; --i) {
            Node* t = q.front(); q.pop();
            t->next = i == 0 ? nullptr : q.front();
            if (t->left) q.push(t->left);
            if (t->right) q.push(t->right);
        }
    }
    return root;
}

// Recursion with implicit O(n) space
Node* connect(Node* root) { // time: O(n); space: O(n)
    if (!root) return root;
    Node* ptr = root->next;
    while (ptr) {
        if (ptr->left) {
            ptr = ptr->left;
            break;
        }
        if (ptr->right) {
            ptr = ptr->right;
            break;
        }
        ptr = ptr->next;
    }
    if (root->right) root->right->next = ptr;
    if (root->left) root->left->next = root->right ? root->right : ptr;
    connect(root->right);
    connect(root->left);
    return root;
}

head代表下一行的開頭第一個node，prev代表下一行我們正要接上next pointer的node。

// Constant Extra Space
Node* connect(Node* root) { // time: O(n); space: O(1)
    if (!root) return root;
    Node *head = nullptr, *prev = nullptr, *cur = root;
    while (cur) {
        // Left child
        if (cur->left) {
            if (prev) {
                prev->next = cur->left;
            } else {
                head = cur->left;
            }
            prev = cur->left;
        }
        // Right child
        if (cur->right) {
            if (prev) {
                prev->next = cur->right;
            } else {
                head = cur->right;
            }
            prev = cur->right;
        }
        cur = cur->next;
        if (!cur) {
            cur = head;
            head = nullptr;
            prev = nullptr;
        }
    }
    return root;
}