脚本宝典收集整理的这篇文章主要介绍了[LeetCode] Populating Next Right Pointers in Each Node，脚本宝典觉得挺不错的，现在分享给大家，也给大家做个参考。
Populating Next Right Pointers in each Node
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.
BFS
Time Complexity
O(N)
Space Complexity
O(N)
思路
Because it is a PErfect binary tree. So we use a queue to keep every level. Initialize a node called PRe, if it is the First node in this level, make pre equals current first node, if it is not, then pre.next = cur. When this level of bfs ends, this level's nodes are all connected.
代码
public void connect(TreeLinkNode root) {
//corner case
if(root == null) return;
Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
queue.offer(root);
TreeLinkNode pre = null;
while(!queue.isEmpty()){
int size = queue.size();
for(int i = 0; i < size; i++){
TreeLinkNode cur = queue.poll();
if(cur.left != null) queue.offer(cur.left);
if(cur.right != null) queue.offer(cur.right);
if(pre != null){
pre.next = cur;
}
pre = cur;
}
pre = null;
}
}
Populating Next Right Pointers in Each Node II
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?
My previous solution still works. :)
以上是脚本宝典为你收集整理的[LeetCode] Populating Next Right Pointers in Each Node全部内容，希望文章能够帮你解决[LeetCode] Populating Next Right Pointers in Each Node所遇到的问题。
本图文内容来源于网友网络收集整理提供，作为学习参考使用，版权属于原作者。
如您有任何意见或建议可联系处理。小编QQ：384754419，请注明来意。