[Leetcode] 116. Populating Next Right Pointers in Each Node 解题报告

题目：

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

思路：

我刚开始想到的思路是层次遍历，对于每一层的元素顺次设置next值即可。然而这样耗费的空间复杂度是O(n)量级的，不符合题目要求。采用递归则可以顺利解决这一问题。对于根节点，我们只需要连接它在每个层次上左子树的最右节点和右子树的最左节点，然后递归调用左子树和右子树即可完成所有next指针值的设置。

代码：

/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
* int val;
* TreeLinkNode *left, *right, *next;
* TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
if(root == NULL) {
return;
}
TreeLinkNode *left_node = root->left;
TreeLinkNode *right_node = root->right;
while(left_node != NULL) {
left_node->next = right_node;
left_node = left_node->right;
right_node = right_node->left;
}
connect(root->left);
connect(root->right);
}
};