116. Populating Next Right Pointers in Each Node 连接右节点

标签：pointer   one   may   use   连接   count   递归   not   figure   

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

思路：traverse的递归。左边的不为空，才能连接到右边
return;表示树中的递归结束了，不需要返回任何东西

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

View Code

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116. Populating Next Right Pointers in Each Node 连接右节点

标签：pointer   one   may   use   连接   count   递归   not   figure   

原文地址：https://www.cnblogs.com/immiao0319/p/13882171.html