116

116. Populating Next Right Pointers in Each Node (Medium)

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 1:

Input: root = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]
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. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.

Example 2:

Input: root = []
Output: []

 

Constraints:

• The number of nodes in the tree is in the range [0, 212 - 1].
• -1000 <= Node.val <= 1000

 

Follow-up:

• You may only use constant extra space.
• The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.

Companies:
Facebook, Microsoft, Amazon, Bloomberg

Related Topics:
Tree, Depth-First Search, Breadth-First Search, Binary Tree

Similar Questions:

• Populating Next Right Pointers in Each Node II (Medium)
• Binary Tree Right Side View (Medium)

Solution 1. BFS

// OJ: https://leetcode.com/problems/populating-next-right-pointers-in-each-node/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    Node* connect(Node* root) {
        if (!root) return nullptr;
        queue<Node*> q{{root}};
        while (q.size()) {
            int cnt = q.size();
            Node *prev = nullptr;
            while (cnt--) {
                auto node = q.front();
                q.pop();
                if (prev) prev->next = node;
                prev = node;
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
        }
        return root;
    }
};

Solution 2. Recursive

The tree is a perfect binary tree. We can connect the nodes layer by layer.

We assume the current layer is already connected, and we iterate through the current layer to connect the nodes in the next layer.

This assumption is true because the root layer is connected, and if we do this layer-by-layer approach, each layer must be already connected when being visited.

// OJ: https://leetcode.com/problems/populating-next-right-pointers-in-each-node/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
public:
    Node* connect(Node* root) {
        if (!root || !root->left) return root;
        auto p = root;
        for (; p; p = p->next) {
            p->left->next = p->right;
            if (p->next) p->right->next = p->next->left;
        }
        connect(root->left);
        return root;
    }
};

Solution 3. Iterative

// OJ: https://leetcode.com/problems/populating-next-right-pointers-in-each-node/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    Node* connect(Node* root) {
        auto first = root;
        while (first && first->left) {
            for (auto p = first; p; p = p->next) {
                p->left->next = p->right;
                if (p->next) p->right->next = p->next->left;
            }
            first = first->left;
        }
        return root;
    }
};