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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
这道题我采用的是广度优先搜索，采用队列来做，同时记下每一行的结点个数
#include<iostream>
#include<queue>
using namespace std;
//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) {}
};
void connect(TreeLinkNode *root) {
queue<TreeLinkNode*> temp_queue;
if(root==NULL)
return;
temp_queue.push(root);
int row_size=1;
while(!temp_queue.empty())
{
TreeLinkNode *p1=NULL;
TreeLinkNode *p2=NULL;
while(row_size>0)
{
if(row_size==1)
{
p1=temp_queue.front();
temp_queue.pop();
p1->next=NULL;
}
else
{
p1=temp_queue.front();
temp_queue.pop();
p2=temp_queue.front();
p1->next=p2;
}
row_size--;
if(p1->left!=NULL)
temp_queue.push(p1->left);
if(p1->right!=NULL)
temp_queue.push(p1->right);
}
row_size=temp_queue.size();
}
}
int main()
{
TreeLinkNode* root=(TreeLinkNode*)malloc(sizeof(TreeLinkNode));
root->val=1;
root->next=NULL;
root->left=(TreeLinkNode*)malloc(sizeof(TreeLinkNode));
root->right=(TreeLinkNode*)malloc(sizeof(TreeLinkNode));
root->left->val=2;
root->right->val=3;
root->left->next=NULL;
root->right->next=NULL;
connect(root);
}
leetcode_116题——Populating Next Right Pointers in Each Node（树，广度优先搜索）
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原文地址：http://www.cnblogs.com/yanliang12138/p/4543481.html
