leetcode----116. Populating Next Right Pointers in Each Node

链接：

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

大意：

给定一棵完全二叉树root，该二叉树的所有叶子节点都在同一层。该二叉树数据结构定义如下：

class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}

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

left和right域的解释与普通二叉树一样。而每个节点的next域指向同一层下一个位于它右边的节点，如果右边没有节点，则指向空，现在需要补全每个节点的next域指向信息。例子：

注意：上述说明中的$id为节点的编号，val才是节点的值

思路：

使用二叉树层次遍历的递归解法。

关于二叉树的层次遍历解法还可以看另一题：使用递归法解决按Z字形打印二叉树节点

代码：

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

    public Node() {}

    public Node(int _val,Node _left,Node _right,Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
/*
// $id是节点的编号 val才是节点的值
{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
*/
// 可用层次遍历的递归法解决
class Solution {
    public Node connect(Node root) {
        connect(root, 0, new ArrayList<>());
        return root;
    }
    public void connect(Node root, int h, List<List<Node>> list) {
        if (root == null)
            return ;
        if (h == list.size())
            list.add(new ArrayList<>());
        if (list.get(h).size() > 0) {
            list.get(h).get(list.get(h).size() - 1).next = root;
        }
        list.get(h).add(root);
        connect(root.left, h + 1, list);
        connect(root.right, h + 1, list);
    }
}

结果：

结论：

理清思路，想好方法，动手写代码。