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Populating Next Right Pointers in Each Node II
Total Accepted: 9695 Total
Submissions: 32965
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
题意：给定一棵随意二叉树（不一定是perfect binary tree），将它每个节点的next指针都指向该节点右边的节点
思路：bfs
这里不能用dfs了，仅仅能用bfs
bfs遍历将同一层的节点存放在同一个数组里，
然后在遍历每一个数组，将前面的节点和后面的节点connect起来，
最后一个节点和NULL connect起来
须要定义一个新的struct结构，保存指向每一个节点的指针和该节点所在的层
复杂度：时间O(n)， 空间O( n)
相关题目：
Populating Next Right Pointers in Each Node
/**
* 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: struct TreeLinkNodeWithLevel
{
TreeLinkNode *p;
int level;
TreeLinkNodeWithLevel(TreeLinkNode *pp, int l):p(pp), level(l){}
}; void connect(TreeLinkNode *root) {
vector<vector<TreeLinkNode *> > nodes;
queue<TreeLinkNodeWithLevel> q;
q.push(TreeLinkNodeWithLevel(root, 0));
while (!q.empty())
{
TreeLinkNodeWithLevel cur = q.front(); q.pop();
if(!cur.p) continue;
if(cur.level >= nodes.size()){
vector<TreeLinkNode *> temp;
temp.push_back(cur.p);
nodes.push_back(temp);
}else{
nodes[cur.level].push_back(cur.p);
}
TreeLinkNodeWithLevel left(cur.p->left, cur.level + 1);
TreeLinkNodeWithLevel right(cur.p->right, cur.level + 1);
q.push(left);
q.push(right);
}
for(int i = 0; i < nodes.size(); i++){
for(int j = 0; j < nodes[i].size() - 1; j++){
nodes[i][j]->next = nodes[i][j + 1];
}
nodes[i][nodes[i].size() - 1]->next = NULL;
}
}
};