116. Populating Next Right Pointers in Each Node.md

116. Populating Next Right Pointers in Each Node

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leetcode-cn Daily Challenge on October 15th, 2020.

leetcode Daily Challenge on November 13th, 2020.

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Difficulty : Medium

Related Topics : Tree、DFS

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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.

**Follow up:

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

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.

Constraints:

• The number of nodes in the given tree is less than 4096.
• -1000 <= node.val <= 1000

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Solution

• mine
  • Java
    • Runtime: 0 ms, faster than 100.00%, Memory Usage: 39 MB, less than 16.78% of Java online submissions

      // O(N)time
      // O(1)space
      public Node connect(Node root) {
          dfs(root, null);
          return root;
      }
      
      void dfs(Node node, Node next){
          if(node == null) return;
          if(node.left != null) {
              node.left.next = node.right;
          }
          if(node.right != null){
              if(next != null) node.right.next = next.left;
          }
          dfs(node.left, node.right);
          dfs(node.right, next == null ? next : next.left);
      
      }
      

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• the most votes

• Runtime: 0 ms, faster than 100.00%, Memory Usage: 39.2 MB, less than 16.78% of Java online submissions

  // O(N)time
  // O(1)space
  public Node connect(Node root) {
      Node level_start=root;
      while(level_start!=null){
          Node cur=level_start;
          while(cur!=null){
              if(cur.left!=null) cur.left.next=cur.right;
              if(cur.right!=null && cur.next!=null) cur.right.next=cur.next.left;
  
              cur=cur.next;
          }
          level_start=level_start.left;
      }
      return root;
  }
  

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