题目描述
Follow up for problem "Populating Next Right Pointers in Each Node".
What if the given tree could be any binary tree? Would your previous solution still work?
Note:
You may only use constant extra space.
For example,
Given the following binary tree,
1
/ \
2 3
/ \ \
4 5 7
After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ \
4-> 5 -> 7 -> NULL
思路
将树与链表结合起来的一道题，在第一层的时候通过left与right将第二层的节点连接起来，在第二层又连接第三层，那么如何将从第一层到第二层呢，用队列空间复杂度不合要求，所以可以考虑使用头节点来连接链表，可以顺理成章的从链表的头结点顺着遍历第二层的节点。
代码
/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
* int val;
* TreeLinkNode left, right, next;
* TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
public void connect(TreeLinkNode root) {
while (root != null) {
//创建一个头结点
TreeLinkNode head = new TreeLinkNode(0);
TreeLinkNode cur = head;
while (root != null) {
//将下一层节点连接成链表
if (root.left != null){
cur.next = root.left;
cur = cur.next;
}
if (root.right != null){
cur.next = root.right;
cur = cur.next;
}
// 继续遍历下一个节点
root = root.next;
}
// 如果遍历完该层，则从下一层的头结点下一个节点开始遍历
root = head.next;
}
}
}
