Q:
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
My solution
/**
* 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 == nullptr) {
return;
}
if(root->left != nullptr)
{
root->left->next = root->right;
}
if(root->right != nullptr && root->next != nullptr)
{
root->right->next = root->next->left;
}
connect(root->left);
connect(root->right);
}
};
这题初看，肯定会用BFS去解决(实际上是可以的),但是其实用DFS会更加方便。该解决方法的核心就是要注意到：root->right->next = root->next->left;。
如果有任何疑问，欢迎留言，或者给我发邮件(liuhiter@gmail.com)。另外无耻求光顾本渣的gayhub leetcode项目。
