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:
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
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
struct TreeLinkNode {
int val;
TreeLinkNode *left, *right, *next;
TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
};
void connect(TreeLinkNode *root) {
queue<pair<TreeLinkNode*,int>>Q;
if (!root)
return;
Q.push(make_pair(root,0));
TreeLinkNode* Levelnode = new TreeLinkNode(0);
int Curlevel = 0;
while (!Q.empty())
{
TreeLinkNode* tmp = Q.front().first;
if (Q.front().second!=Curlevel){
Levelnode->next = NULL;
Levelnode = tmp;
Curlevel++;
}
else{
Levelnode ->next = tmp;
Levelnode = Levelnode->next;
}
Q.pop();
if (tmp->left)
Q.push(make_pair(tmp->left, Curlevel + 1));
if (tmp->right)
Q.push(make_pair(tmp->right, Curlevel + 1));
}
}
Populating Next Right Pointers in Each Node
原文地址:http://blog.csdn.net/li_chihang/article/details/44747809