《中英双解》leetCode Populating Next Right Pointers in Each Node

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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.

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.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

这第一种方法就是用二叉树的层序遍历 

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}
    
    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _left, Node _right, Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/

class Solution {
   public Node connect(Node root) {
      if(root == null){
         return null;
      }
      Deque<Node> queue = new LinkedList<>();
      queue.add(root);
      while(!queue.isEmpty()){
         int size = queue.size();
         for(int i = 0;i < size;i++){
            Node node = queur.poll();
            if(i < size - 1){
               node.next = queue.peak();
            }
            if(node.left != null){
               queue.add(node.left);
            }
            if(node.right != null){
               queue.add(node.right);
            }
         }
      }
      return root;
   }
}

 下一个方法比较巧妙，注意next节点的使用

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}
    
    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _left, Node _right, Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/

class Solution {
    public Node connect(Node root) {
       if(root == null){
          return null;
       }
       if(root.left != null){
          root.left.next = root.right;
          if(root.next != null){
             root.right.next = root.next.left;
          }
       }
       connect(root.left);
       connect(root.right);
       return root;
    }
}

顺便复习一下二叉树的层序遍历

class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
       if(root == null){
          return new ArrayList<>();
       }
       List<List<Integer>> res = new ArrayList<>();
       Queue<TreeNode> queue = new LinkedList<>();
       queue.add(root);
       while(!queue.isEmpty()){
          int size = queue.size();
          List<Integer> list  = new ArrayList<>();
          for(int i = 0;i < size;i++){
             TreeNode node = queue.poll();
             list.add(node.val);
             if(node.left != null) queue.add(node.left);
             if(node.right != null) queue.add(node.right);
          }
          res.add(list);
       }
       return res;
    }
}

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