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
/**
* 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;
root->next = NULL;
// 先右后左，切记
connect(root, root->right, 1);
connect(root, root->left, 0);
}
void connect(TreeLinkNode *root, TreeLinkNode *child, int flag)
{
if (child == NULL)
return;
if (flag == 0)
{
if (root->right != NULL)
child->next = root->right;
else
{
TreeLinkNode *temp = root->next;
while (temp != NULL && temp->left == NULL && temp->right == NULL)
{
temp = temp->next;
}
if (temp == NULL)
child->next = NULL;
else if (temp->left != NULL)
child->next = temp->left;
else
child->next = temp->right;
}
connect(child, child->right, 1);
connect(child, child->left, 0);
} else
{
if (root->next == NULL)
child->next = NULL;
else
{
TreeLinkNode *temp = root->next;
while (temp != NULL && temp->left == NULL && temp->right == NULL)
{
temp = temp->next;
}
if (temp == NULL)
child->next = NULL;
else if (temp->left != NULL)
child->next = temp->left;
else
child->next = temp->right;
}
connect(child, child->right, 1);
connect(child, child->left, 0);
}
}
};