Populating Next Right Pointers in Each Node
Given a binary treestruct 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
Hide Tags
Tree Depth-first Search
Hide Similar Problems
(H) Populating Next Right Pointers in Each Node II (M) Binary Tree Right Side View
一边 DFS,一边利用当前节点的next,来构造子节点的next即可。
/**
* 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 && root->left)
{
TreeLinkNode* levelHead=root;
TreeLinkNode* p = levelHead;
root->next = NULL;
while(p->left)
{
p->left->next = p->right;
if(NULL != p->next)
{
p->right->next = p->next->left;
p = p->next;
}
else
{
p->right->next = NULL;
levelHead = levelHead->left;
p = levelHead;
}
}
}
return;
}
};
24 ms.
No comments:
Post a Comment