LeetCode题解：Populating Next Right Pointers in Each Node

Given a binary tree

struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

You may only use constant extra space.

You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,

Given the following perfect binary tree,

1

/ \

2 3

/ \ / \

4 5 6 7

After calling your function, the tree should look like:

1 -> NULL

/ \

2 -> 3 -> NULL

/ \ / \

4->5->6->7 -> NULL

题意：让给定二叉树的每一层的结点从左到右指向它的右结点，最右边的结点则指向NULL

解决思路：层序遍历

代码：

public class Solution {
public void connect(TreeLinkNode root) {
while (root != null) {
TreeLinkNode curr = root;

while (curr != null && curr.left != null) {
curr.left.next = curr.right;
curr.right.next = curr.next == null ? null : curr.next.left;
curr = curr.next;
}

root = root.left;
}
}
}