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116.填充每个节点的下一个右侧节点指针.java
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116.填充每个节点的下一个右侧节点指针.java
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/*
* @lc app=leetcode.cn id=116 lang=java
*
* [116] 填充每个节点的下一个右侧节点指针
*
* https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node/description/
*
* algorithms
* Medium (44.11%)
* Likes: 86
* Dislikes: 0
* Total Accepted: 13.3K
* Total Submissions: 28.7K
* Testcase Example: '{"$id":"1","left":{"$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":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}'
*
* 给定一个完美二叉树,其所有叶子节点都在同一层,每个父节点都有两个子节点。二叉树定义如下:
*
* struct Node {
* int val;
* Node *left;
* Node *right;
* Node *next;
* }
*
* 填充它的每个 next 指针,让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点,则将 next 指针设置为 NULL。
*
* 初始状态下,所有 next 指针都被设置为 NULL。
*
*
*
* 示例:
*
*
*
*
* 输入:{"$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。
*
*
*
*
* 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。
*
*
*/
/*
// 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 root;
}
if (root.left != null) {
root.left.next = root.right;
Node left = root.left;
Node right = root.right;
while (left != null) {
left.next = right;
left = left.right;
right = right.left;
}
}
connect(root.left);
connect(root.right);
return root;
}
}