leetcode116. 填充每个节点的下一个右侧节点指针

旧城等待， 2023-07-06 11:38 1阅读 0赞

[116. 填充每个节点的下一个右侧节点指针][116.]

难度中等128

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：**

![format_png][]

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。

思路：层序遍历稍微改一下，把相同层的连一下即可。

/*
    // 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;
            }
    
            Queue<Node> Q = new LinkedList<Node>(); 
            Q.add(root);
            while (Q.size() > 0) {
                int size = Q.size();
                for(int i = 0; i < size; i++) {
                    Node node = Q.poll();
                    if (i < size - 1) {
                        node.next = Q.peek();
                    }
                    if (node.left != null) {
                        Q.add(node.left);
                    }
                    if (node.right != null) {
                        Q.add(node.right);
                    }
                }
            }
            return root;
        }
    }

[116.]: https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node/
[format_png]: https://imgconvert.csdnimg.cn/aHR0cHM6Ly9hc3NldHMubGVldGNvZGUtY24uY29tL2FsaXl1bi1sYy11cGxvYWQvdXBsb2Fkcy8yMDE5LzAyLzE1LzExNl9zYW1wbGUucG5n?x-oss-process=image/format,png