LeetCode 填充每个节点的下一个右侧节点指针（java）

最新推荐文章于 2020-10-17 13:33:55 发布

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最新推荐文章于 2020-10-17 13:33:55 发布

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分类专栏： LeetCode 文章标签：数据结构二叉树

本文链接：https://blog.youkuaiyun.com/qq_37174705/article/details/88867193

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本文介绍了一种算法，用于填充完美二叉树中每个节点的Next指针，使其指向右侧相邻节点。通过层级遍历实现，确保了每个节点的Next指针正确设置。

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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。

思路：

这就是二叉树的层级遍历的变种

利用一个int变量来标记每层的大小，再结合层级遍历

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;
    }
};
*/
class Solution {
    public Node connect(Node root) {
        if(root == null){
            return null;
        }
        java.util.LinkedList<Node> q = new java.util.LinkedList<>();
        q.add(root);
        int count = 1;//用来标记每层有多少个节点
        while(q.size() != 0){
            Node node1 = q.removeFirst();
            count--;
            if(node1.left != null){
                q.add(node1.left);
            }
            if(node1.right != null){
                q.add(node1.right);
            }   
            //count=0，表示该层已经遍历结束
            if(count == 0){
               node1.next = null;
               count = q.size();
            }else{
               node1.next = q.getFirst(); 
            }
        }
        return root;
    }
}