填充每个节点的下一个右侧节点指针

填充每个节点的下一个右侧节点指针

题目：给定一个完美二叉树，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：

struct Node {
int val;
Node *left;
Node *right;
Node *next;
}
填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。

初始状态下，所有 next 指针都被设置为 NULL。

输入：{"$id":"1","left":{"$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}

输出：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}

解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。

解题思路：运用递归来解决，如果当前节点是父节点的左节点则将next设置为父节点的右节点，如果是右节点则将父节点的next节点的左节点设置为next即可

/*
// 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;
        dfs(root, null);
        return root;
    }
    
    /**
    需要借助父节点的next来寻找next
    **/
    private void dfs(Node cur, Node parent) {
        if(cur == null)
            return ;
        
        //root
        if(parent == null) {
            cur.next = null;
            dfs(cur.left, cur);
            dfs(cur.right, cur);
            return ;
        }
            
        if(cur == parent.left) {
            cur.next = parent.right;
        } else {
            cur.next = parent.next == null ? null : parent.next.left;
        }
        
        dfs(cur.left, cur);
        dfs(cur.right, cur);
    }
}