117. Populating Next Right Pointers in Each Node II.md

117. Populating Next Right Pointers in Each Node II

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leetcode-cn Daily Challenge on September 28th, 2020.

leetcode Daily Challenge on December 6th, 2020.

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Difficulty : Medium

Related Topics : Tree、DFS

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Given a binary tree

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,null,7]
Output: [1,#,2,3,#,4,5,7,#]
Explanation: Given the above 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 6000.
• -100 <= node.val <= 100

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Solution

• mine
  • Java
    • Runtime: 1 ms, faster than 59.89%, Memory Usage: 38.9 MB, less than 90.79% of Java online submissions

      // O(N)time
      // O(D)space
      Map<Integer, Node> map = new HashMap<>();
      public Node connect(Node root) {
          helper(root, 0);
          return root;
      }
      
      void helper(Node node, int deepth){
          if(node == null) return;
          if(map.containsKey(deepth)){
              map.get(deepth).next = node;
          }
          map.put(deepth, node);
          helper(node.left, deepth + 1);
          helper(node.right, deepth + 1);
      }
      

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• the most votes

• Runtime: 0 ms, faster than 100.00%, Memory Usage: 38.6 MB, less than 99.34% of Java online submissions

  // O(N)time
  // O(1)space
  public Node connect(Node root) {
      Node head = null; //head of the next level
      Node prev = null; //the leading node on the next level
      Node cur = root;  //current node of current level
  
      while (cur != null) {
          while (cur != null) { //iterate on the current level
              //left child
              if (cur.left != null) {
                  if (prev != null) {
                      prev.next = cur.left;
                  } else {
                      head = cur.left;
                  }
                  prev = cur.left;
              }
              //right child
              if (cur.right != null) {
                  if (prev != null) {
                      prev.next = cur.right;
                  } else {
                      head = cur.right;
                  }
                  prev = cur.right;
              }
              //move to next node
              cur = cur.next;
          }
  
          //move to next level
          cur = head;
          head = null;
          prev = null;
      }
      return root;
  }
  

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