notes/OJ notes/pages/Leetcode Populating-Next-Right-Pointers-In-Each-Node.md

Leetcode Populating-Next-Right-Pointers-In-Each-Node

2022-07-16 09:43

Algorithms:

#algorithm #BFS #optimization #DFS

Data structures:

#DS #binary_tree

Difficulty:

#coding_problem #difficulty-medium

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

tag:#BFS

Links:

• Link to problem

---

Problem

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.

Examples

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

Constraints:

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

Thoughts

[!summary] This is a #BFS problem, and cam be optimized

BFS way of doing this, O(n)time, O(n + 1)space

simple BFS, for each level, connect the node to the next.

Remember to get the for loop right.

BFS-like iteration, two pointers

We utilize the ->next property to link ptr->left->right->next to ptr->right->left

Also can be done with DFS

Solution

BFS unoptimized:

/*
// 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) {
        // BFS.
        // Can we do it with DFS?
        queue<Node*> todo;
        
        if (root && root->left && root->right) {
            todo.push(root->left);
            todo.push(root->right);
        }
        
        Node *ptr, *nextPtr;
        
        while (!todo.empty()) {
            nextPtr = todo.front();
            todo.pop();
            for (size_t i = 0, size = todo.size(); i < size; i++) {
                ptr = nextPtr;
                nextPtr = todo.front();
                todo.pop();
                
                ptr->next = nextPtr;
                if (ptr->left) {
                    todo.push(ptr->left);
                    todo.push(ptr->right);
                }
            }
            if (nextPtr->left) {
                todo.push(nextPtr->left);
                todo.push(nextPtr->right);
            }
        }
        
        return root;
    }
};