logseq_notes/pages/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_problems #difficulty_medium

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

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:**
  
  ![](https://assets.leetcode.com/uploads/2019/02/14/116_sample.png)
  
  **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 It relies on the fact that it is perfect binary tree. The core part for optimization is utilizing ptr->next

Using ptr->next in DFS and iteration enable us to traverse between branches like in BFS

!

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, O(n)time, O(n + 1)space

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

!

We can't use ptr->left->right->next = ptr->right->left, since this is only one level, and doesn't work on two levels:

!

Also can be done with DFS O(n), O(n)

We recursively fill next pointers the level below current.

Base case:

• Root->left is empty, which means this is the leaf level, quit.

  Pseudo-code:

  1. Check for base case
  2. link root->left and root->right
  3. link root->right and root->next->left (if root->next is valid.)

Solution

DFS:

/*
// 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 {
void conn(Node *root) {
  // preorder DFS, here we go!
  // It assumes at least one level, which doesn't include []
  if (root->left) {
    root->left->next = root->right;
    if (root->next) {
      root->right->next = root->next->left;
    }
    conn(root->left);
    conn(root->right);
  }
}

public:
Node *connect(Node *root) {
  if (root) {
    conn(root);
  }
  return root;
}
};

Iterative, O(n)time O(1)space

class Solution {
public:
Node *connect(Node *root) {
  // iteration
  Node *ptr = root;
  Node *nextPtr = nullptr;
  while (ptr && ptr->left) {
    ptr->left->next = ptr->right;
    if (!nextPtr) {
      nextPtr = ptr->left;
    }
    if (!ptr->next) {
      ptr = nextPtr;
      nextPtr = nullptr;
    } else {
      ptr->right->next = ptr->next->left;
      ptr = ptr->next;
    }
  }
  return root;
}
};

BFS unoptimized:

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;
}
};