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链接：

大意：

给定一棵完全二叉树root，该二叉树的所有叶子节点都在同一层。该二叉树数据结构定义如下：

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;    }};

left和right域的解释与普通二叉树一样。而每个节点的next域指向同一层下一个位于它右边的节点，如果右边没有节点，则指向空，现在需要补全每个节点的next域指向信息。例子：

注意：上述说明中的$id为节点的编号，val才是节点的值

思路：

使用二叉树层次遍历的递归解法。

关于二叉树的层次遍历解法还可以看另一题：

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;    }};/*// $id是节点的编号 val才是节点的值{"$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}*/// 可用层次遍历的递归法解决class Solution {    public Node connect(Node root) {        connect(root, 0, new ArrayList<>());        return root;    }    public void connect(Node root, int h, List
   
    
     > list) {        if (root == null)            return ;        if (h == list.size())            list.add(new ArrayList<>());        if (list.get(h).size() > 0) {            list.get(h).get(list.get(h).size() - 1).next = root;        }        list.get(h).add(root);        connect(root.left, h + 1, list);        connect(root.right, h + 1, list);    }}
    
   

结果：

结论：

理清思路，想好方法，动手写代码。 

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