LeetCode116、填充每个节点的下一个右侧节点

题目描述

https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node/

解法:

/*
// 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;
        //从所给的图上看，链接的是同一层上的结点，利用递归我们可以把问题拆分：
        Node left = connect(root.left);
        Node right = connect(root.right);

        //链接的做法：
        //同一父节点的兄弟结点
        if(root.left!=null){
    
    
             root.left.next = root.right;
       // root.right.next = null;
        }
        //不同父节点的非兄弟结点相连，left和right都是root的孩子
        while(left!=null&&left.right!=null){
    
    
              left.right.next = right.left;
              left = left.right;
              right = right.left;//left和righ必然同时空，完美二叉树
        }
        
       return root;

        
    }
}

在书：labuladong的算法小抄更为清晰的做法是：