标签：zigzag des roo 层次遍历 atoi NPU script dea char
作者:
晚于: 2020-08-05 12:00:00后提交分数乘系数50%
截止日期: 2020-08-12 12:00:00
问题描述 :
给定一个二叉树，返回其节点值的锯齿形层次遍历。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。
例如：
给定二叉树 [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回锯齿形层次遍历如下：
[
[3],
[20,9],
[15,7]
]
程序输出：
3 20 9 15 7
可使用以下main函数：
#include <iostream>
#include <queue>
#include <cstdlib>
#include <cstring>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(NULL), right(NULL) {}
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
TreeNode* inputTree()
{
int n,count=0;
char item[100];
cin>>n;
if (n==0)
return NULL;
cin>>item;
TreeNode* root = new TreeNode(atoi(item));
count++;
queue<TreeNode*> nodeQueue;
nodeQueue.push(root);
while (count<n)
{
TreeNode* node = nodeQueue.front();
nodeQueue.pop();
cin>>item;
count++;
if (strcmp(item,"null")!=0)
{
int leftNumber = atoi(item);
node->left = new TreeNode(leftNumber);
nodeQueue.push(node->left);
}
if (count==n)
break;
cin>>item;
count++;
if (strcmp(item,"null")!=0)
{
int rightNumber = atoi(item);
node->right = new TreeNode(rightNumber);
nodeQueue.push(node->right);
}
}
return root;
}
int main()
{
TreeNode* root;
root=inputTree();
vector<vector<int> > res=Solution().zigzagLevelOrder(root);
for(int i=0; i<res.size(); i++)
{
vector<int> v=res[i];
for(int j=0; j<v.size(); j++)
cout<<v[j]<<" ";
}
}
输入说明 :
首先输入结点的数目n（注意，这里的结点包括题中的null空结点）
然后输入n个结点的数据，需要填充为空的结点，输入null。
输出说明 :
输出结果，每个数据的后面跟一个空格。
输入范例 :
输出范例 :
#include <iostream> #include <queue> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode() : val(0), left(NULL), right(NULL) {} TreeNode(int x) : val(x), left(NULL), right(NULL) {} TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} }; class Solution { public: vector<vector<int>> zigzagLevelOrder(TreeNode* root) { vector<vector<int>> res; if(!root) return res; queue<TreeNode*> Q; Q.push(root); int m=0; while(!Q.empty()) { m++; vector<int> temp; int len=Q.size(); for(int i=0;i<len;i++) { temp.push_back(Q.front()->val); if(Q.front()->left) Q.push(Q.front()->left); if(Q.front()->right) Q.push(Q.front()->right); Q.pop(); } if(m%2!=0) res.push_back(temp); else { reverse(temp.begin(),temp.end()); res.push_back(temp); } } return res; } }; TreeNode* inputTree() { int n,count=0; char item[100]; cin>>n; if (n==0) return NULL; cin>>item; TreeNode* root = new TreeNode(atoi(item)); count++; queue<TreeNode*> nodeQueue; nodeQueue.push(root); while (count<n) { TreeNode* node = nodeQueue.front(); nodeQueue.pop(); cin>>item; count++; if (strcmp(item,"null")!=0) { int leftNumber = atoi(item); node->left = new TreeNode(leftNumber); nodeQueue.push(node->left); } if (count==n) break; cin>>item; count++; if (strcmp(item,"null")!=0) { int rightNumber = atoi(item); node->right = new TreeNode(rightNumber); nodeQueue.push(node->right); } } return root; } int main() { TreeNode* root; root=inputTree(); vector<vector<int> > res=Solution().zigzagLevelOrder(root); for(int i=0; i<res.size(); i++) { vector<int> v=res[i]; for(int j=0; j<v.size(); j++) cout<<v[j]<<" "; } }
标签：zigzag des roo 层次遍历 atoi NPU script dea char
原文地址：https://www.cnblogs.com/zmmm/p/13632801.html
