116. 填充每个节点的下一个右侧节点指针

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。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

模拟队列

class Solution {
    public Node connect(Node root) {
        if (root == null) {
            return null;
        }
        Node cur = root, tail = root;
        int size = 1;
        while (cur != null) {
            Node pre = null;
            for (int i = 1; i <= size; ++i) {
                if (cur.left != null) {
                    tail.next = cur.left;
                    tail = cur.left;
                }
                if (cur.right != null) {
                    tail.next = cur.right;
                    tail = cur.right;
                }
                if (pre != null) {
                    pre.next = cur;
                }
                pre = cur;
                cur = cur.next;
            }
            pre.next = null;
            size <<= 1;
        }
        return root;
    }
}

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;
    }
}

官方的题解

class Solution {
    public Node connect(Node root) {
        if (root == null) {
            return root;
        }

        // 从根节点开始
        Node leftmost = root;

        while (leftmost.left != null) {

            // 遍历这一层节点组织成的链表，为下一层的节点更新 next 指针
            Node head = leftmost;

            while (head != null) {

                // CONNECTION 1
                head.left.next = head.right;

                // CONNECTION 2
                if (head.next != null) {
                    head.right.next = head.next.left;
                }

                // 指针向后移动
                head = head.next;
            }

            // 去下一层的最左的节点
            leftmost = leftmost.left;
        }

        return root;
    }
}

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;
    }
}