70：Populating Next Right Pointers in Each Node II

题目：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

解析一：我们最先想到的应该是借助于层序遍历，但其要求的空间为 O(log n)，不符合题目中对常数空间的要求，不过该方法的代码如下：

//迭代版本，时间复杂度 O(n)，空间复杂度 O(log n)
class Solution {
public:
void connect(TreeLinkNode* root) {
queue<TreeLinkNode*> q1, q2;
if (root) q1.push(root)

while (true) {
if (q1.empty()) break;
while (!q1.empty()) {
auto p = q1.front(), q1.pop();

if (!q1.empty())
p -> next = q1.front();
else p -> next = nullptr;

if (p -> left != nullptr) q2.push(p -> left);
if (p -> right != nullptr) q2.push(p -> right);
}
while (!q2.empty())
q1.push(q2.front()), q2.pop();
}
}
};

解析二：常数空间的代码分为递归和迭代两种版本，不过它们的思想是一模一样的，并且此递归为尾递归

下面解法代码的思想及编写参考了网址https://github.com/soulmachine/leetcode#leetcode题解题目

递归代码如下：

// 递归版本，时间复杂度 O(n)，空间复杂度 O(1)
class Solution {
public:
void connect(TreeLinkNode* root) {
if (!root) return;

TreeLinkNode dummy{-1};
for (TreeLinkNode *curr = root, *prev = &dummy; curr; curr = curr -> next) {
if (curr -> left != nullptr) {
prev -> next = cur -> left;
prev = prev -> next;
}
if (cur -> right != nullptr) {
prev -> next = curr -> right;
prev = prev -> next;
}
}
connect(dummy.next);
}
};

迭代代码如下：

// 迭代版本，时间复杂度 O(n)，空间复杂度 O(1)
class Solution {
public:
void connect(TreeLinkNode* root) {
TreeLinkNode* cur = root;

while (cur) {
TreeLinkNode dummy{-1};
TreeLinkNode* prev = &dummy;
while (cur) {
if (cur -> left) {
prev -> next = cur -> left;
prev = prev -> next;
}
if (cur -> right) {
prev -> next = cur -> right;
prev = prev -> next;
}
cur = cur -> next;
}
cur = dummy.next;
}
}
};