剑指Offer--060-把二叉树打印成多行

链接

牛客OJ：把二叉树打印成多行

九度OJ：未收录

GitHub代码： 060-把二叉树打印成多行

CSDN题解：剑指Offer–060-把二叉树打印成多行

牛客OJ	九度OJ	CSDN题解	GitHub代码
060-把二叉树打印成多行	未收录	剑指Offer–060-把二叉树打印成多行	060-把二叉树打印成多行

题意

题目描述

请实现一个函数，用来判断一颗二叉树是不是对称的。

注意，如果一个二叉树同此二叉树的镜像是同样的，定义其为对称的

分析

其实就是层次遍历，这个我们在我之前的一篇博客里面将的很清楚了

二叉树的遍历详解（前序中序后序层次-递归和非递归）

剑指Offer–023-从上往下打印二叉树（层次遍历二叉树）

其中调试的时候使用了

剑指Offer–006-重构二叉树来辅助我们通过前序和中序遍历的序列来生成二叉树

直接贴代码，只需要直接将我们层次遍历输出过程，变成将输入压入vector中即可

代码

#include <iostream>
#include <vector>
#include <deque>
#include <queue>

using namespace std;

#define __tmain main

#ifdef __tmain

#define debug cout

#else

#define debug 0 && cout

#endif // __tmain

#define undebug 0 && cout

#ifdef __tmain
struct TreeNode
{
int val;
struct TreeNode *left;
struct TreeNode *right;

TreeNode(int x = 0)
:val(x), left(NULL), right(NULL)
{
}
};
#endif // __tmain

class Solution
{
vector< vector<int> >   res;
vector<int>             curr;
public:
vector< vector<int> > Print(TreeNode *root)
{
LevelOrder(root);

LevelOrderDev(root);

LevelOrderUsePoint(root);

LevelOrderUseSize(root);

LevelOrderUseEnd(root);
return this->res;
}

int PrintLevel(TreeNode *root, int level)
{
if(root == NULL || level < 0)
{
return 0;
}
else if(level == 0)
{
debug <<root->val;
curr.push_back(root->val);        /// add for currult in vector

return 1;
}
else
{
return PrintLevel(root->left, level - 1) + PrintLevel(root->right, level - 1);
}
}

void LevelOrder(TreeNode *root)
{
curr.clear( );           /// add for currult in vector
res.clear( );
if(root == NULL)
{
return;
}
for(int level = 0; ; level++)
{
if(PrintLevel(root, level) == 0)
{
break;
}
res.push_back(curr);
curr.clear( );
debug <<endl;
}
}

//////////////////////////
//////////////////////////
void LevelOrderDev(TreeNode *root)
{
curr.clear( );           /// add for currult in vector
res.clear( );

if(root == NULL)
{
return ;
}
deque<TreeNode *> qFirst, qSecond;
qFirst.push_back(root);

while(qFirst.empty( ) != true)
{
while (qFirst.empty( ) != true)
{
TreeNode *temp = qFirst.front( );
qFirst.pop_front( );

debug << temp->val;
curr.push_back(temp->val);        /// add for currult in vector

if (temp->left != NULL)
{
qSecond.push_back(temp->left);
}
if (temp->right != NULL)
{
qSecond.push_back(temp->right);
}
}
debug << endl;
res.push_back(curr);
curr.clear( );

qFirst.swap(qSecond);

}
}

//////////////////////////
//////////////////////////
void LevelOrderUsePoint(TreeNode *root)
{
curr.clear( );           /// add for currult in vector
res.clear( );
if(root == NULL)
{
return ;
}

vector<TreeNode*> vec;
vec.push_back(root);

int cur = 0;
int end = 1;

while (cur < vec.size())
{
end = vec.size();

while (cur < end)
{
debug << vec[cur]->val;
curr.push_back(vec[cur]->val);        /// add for currult in vector

if (vec[cur]->left != NULL)
{
vec.push_back(vec[cur]->left);

}
if (vec[cur]->right != NULL)
{
vec.push_back(vec[cur]->right);
}
cur++;
}
res.push_back(curr);
curr.clear( );
debug << endl;
}
}

void LevelOrderUseSize(TreeNode *root)
{
curr.clear( );           /// add for currult in vector
res.clear( );
if(root == NULL)
{
return ;
}

int parentSize = 1, childSize = 0;
TreeNode *temp = NULL;

queue<TreeNode *> q;
q.push(root);
while(q.empty( ) != true)
{
temp = q.front( );
debug <<temp->val;
curr.push_back(temp->val);        /// add for currult in vector

q.pop( );

if (temp->left != NULL)
{
q.push(temp->left);

childSize++;
}
if (temp->right != NULL)
{
q.push(temp->right);
childSize++;
}

parentSize--;
if (parentSize == 0)
{
parentSize = childSize;
childSize = 0;
res.push_back(curr);
curr.clear( );
debug << endl;
}

}
}

void LevelOrderUseEnd(TreeNode *root)
{
curr.clear( );           /// add for currult in vector
res.clear( );
if(root == NULL)
{
return ;
}

queue<TreeNode *> q;

q.push(root);
q.push(NULL);

while(q.empty( ) != true)
{
TreeNode* node = q.front();
q.pop();

if (node)
{
debug <<node->val;
curr.push_back(node->val);        /// add for currult in vector

if (node->left != NULL)
{
q.push(node->left);
}
if (node->right != NULL)
{
q.push(node->right);
}
}
else if (q.empty( ) != true)
{
q.push(NULL);

res.push_back(curr);
curr.clear( );
debug << endl;
}
}
if(curr.size() != 0)
{
res.push_back(curr);
curr.clear( );
cout <<endl;
}
}
};

int __tmain( )
{

TreeNode tree[7];

tree[0].val = 8;
tree[0].left = &tree[1];
tree[0].right = &tree[2];

tree[1].val = 6;
tree[1].left = &tree[3];
tree[1].right = &tree[4];

tree[2].val = 6;
tree[2].left = &tree[5];
tree[2].right = &tree[6];

tree[3].val = 5;
tree[3].left = NULL;
tree[3].right = NULL;

tree[4].val = 7;
tree[4].left = NULL;
tree[4].right = NULL;

tree[5].val = 7;
tree[5].left = NULL;
tree[5].right = NULL;

tree[6].val = 5;
tree[6].left = NULL;
tree[6].right = NULL;

Solution solu;
vector< vector<int> > res = solu.Print(tree);
for(int i = 0; i < res.size( ); i++)
{
for(int j = 0; j < res[i].size( ); j++)
{
cout <<res[i][j];
}
cout <<endl;
}

return 0;
}