原题
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。
示例：
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。
提示：
+ 你只能使用常量级额外空间。
+ 使用递归解题也符合要求，本题中递归程序占用的栈空间不算做额外的空间复杂度。
解法
思想
DFS搜索，用一个list存放每层前一个遍历的节点，再次遍历到该层的时候从list中取出上一个元素并修改next指针。
代码
class Solution {
List<Node> list;
public Node connect(Node root) {
list = new ArrayList<Node>();
dfs(root,0);
return root;
}
public void dfs(Node root,int depth){
if(root == null) return;
if(list.size()>depth){
Node node = list.get(depth);
node.next = root;
list.set(depth,root);
}
else list.add(root);
dfs(root.left,depth+1);
dfs(root.right,depth+1);
}
}
原创文章，作者：彭晨涛，如若转载，请注明出处：https://www.codetool.top/article/leetcode116-%e5%a1%ab%e5%85%85%e6%af%8f%e4%b8%aa%e8%8a%82%e7%82%b9%e7%9a%84%e4%b8%8b%e4%b8%80%e4%b8%aa%e5%8f%b3%e4%be%a7%e8%8a%82%e7%82%b9%e6%8c%87%e9%92%88/
