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
Key: consider the case when the next pointer has no child. Must find the following next pointers until one has a child. Populate from the right.
/**
* 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) {
// Note: The Solution object is instantiated only once and is reused by each test case.
if(root==NULL || (root->left==NULL && root->right==NULL))
return;
if(root->left!=NULL && root->right!=NULL)
root->left->next=root->right;
TreeLinkNode *rootLastChild;
if(root->right!=NULL)
rootLastChild=root->right;
else
rootLastChild=root->left;
TreeLinkNode *nextChild=NULL;
TreeLinkNode *rootNext=root->next;
while(rootNext!=NULL)
{
if(rootNext->left!=NULL)
nextChild=rootNext->left;
else if(rootNext->right!=NULL)
nextChild=rootNext->right;
if(nextChild != NULL)
break;
rootNext=rootNext->next;
}
if(nextChild != NULL)
{
rootLastChild->next=nextChild;
}
connect(root->right);
connect(root->left);
}
};