116. Populating Next Right Pointers in Each Node

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

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

Explanation: Given the above perfect 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.

利用Level Order Traversal去掃描每一層，把同一層內的node相接，最後一個node的next指向nullptr。

// 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;
}

當前節點的left node的next node可以跟當前節點的right node接上。 如果當前節點有next node，則當前節點的right node的next node可以跟當前節點的next node的left node接上。

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

// O(1) Extra Space
Node* connect(Node* root) { // time: O(n); space: O(1)
    if (!root) return root;
    Node *start = root, *cur = nullptr;
    while (start->left) {
        cur = start;
        while (cur) {
            cur->left->next = cur->right;
            if (cur->next) cur->right->next = cur->next->left;
            cur = cur->next;
        }
        start = start->left;
    }
    return root;
}