力扣 116 构建树的每一个节点的next节点

题意：给一个树，构建它的每一个节点的next节点

思路：按层遍历树，

1. 每次遍历到一个节点cur时，判断它的pre节点是否为空，如果不为空，那么pre.next = cur,
2. 如果是当前层的最后一个节点那么它的next为null

思想：按层遍历

复杂度：时间O（n)，空间O(n)

/*
// 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();
        q.add(root);
        Node pre = null;
        int level = 1;
        int start = 0;
        int end = 1;
        while(!q.isEmpty()) {
            Node cur = q.poll();
            start++;
            if(pre != null) {
                pre.next = cur;
            }
            pre = cur;
            if(cur.left != null) {
                q.add(cur.left);
                end++;
            }
            if(cur.right != null) {
                q.add(cur.right);
                end++;
            }
            if(level == start) {
                pre = null;
                level = end;
            }
        }        
        return root;
    }
}