116. Populating Next Right Pointers in Each Node
题目
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
解析
- 使用层次遍历，每一层从左到右串接起来就行，每层最后一个元素next置NULL即可！
// Populating Next Right Pointers in Each Node
class Solution_116 {
public:
//运行时间：8ms
//占用内存：892k
void connect(TreeLinkNode *root) {
if (!root)
{
return;
}
queue<TreeLinkNode*> que;
que.push(root);
while (!que.empty())
{
int size = que.size(); //每一层的大小
TreeLinkNode* cur, *pre=NULL;
while (size--)
{
cur= que.front();
que.pop();
if (pre)
{
pre->next = cur;
}
pre = cur;
if (cur->left)
{
que.push(cur->left);
}
if (cur->right)
{
que.push(cur->right);
}
}
cur->next = NULL;
}
}
void connect1(TreeLinkNode *root) {
queue<TreeLinkNode*> q;
if (!root)
return;
q.push(root);
while (!q.empty()) {
int size = q.size();
for (int i = 0; i < size; i++) {
TreeLinkNode* node = q.front();
q.pop();
if (i == size - 1)
node->next = nullptr;
else
node->next = q.front();//当前节点已经出栈， q.front()为下一节点，避免记录上一次节点
if (node->left)
q.push(node->left);
if (node->right)
q.push(node->right);
}
}
}
};
题目来源
