leetcode116-填充每个节点的下一个右侧节点指针

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:

leetcode116-填充每个节点的下一个右侧节点指针

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/