二叉树基础oj练习
1.单值二叉树
题目:
力扣
https://leetcode.cn/problems/univalued-binary-tree/
思路:
单值二叉树 = root和左右孩子的值相等 + 左子树是单值二叉树 + 右子树是单值二叉树
代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
bool isUnivalTree(struct TreeNode* root)
{
if(root==NULL)
return true;
if(root->left&&root->left->val!=root->val)
return false;
if(root->right&&root->right->val!=root->val)
return false;
return isUnivalTree(root->left)&&isUnivalTree(root->right);
}2.相同的树
题目:
力扣https://leetcode.cn/problems/same-tree/submissions/
思路:
相同二叉树 = 比较两棵树的根 + 递归左子树（是否是相同二叉树）+ 递归右子树（是否是相同二叉树）
代码:
bool isSameTree(struct TreeNode* p, struct TreeNode* q)
{
//根,都为空
if(p==NULL&&q==NULL)
return true;
//其中一个根为空
if(p==NULL||q==NULL)
return false;
//都不为空
if(p->val!=q->val)
return false;
return isSameTree(p->left,q->left)&&isSameTree(p->right,q->right);
}3.对称二叉树
题目:
力扣https://leetcode.cn/problems/symmetric-tree/
思路:
解题方法和上一道题相似
原来的函数接口上是递归不起来的，因此要写一个子函数。
代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
bool _isSymmetricTree(struct TreeNode* root1,struct TreeNode*root2)
{
if(root1==NULL&&root2==NULL)
return true;
if(root1==NULL||root2==NULL)
return false;
if(root1->val!=root2->val)
return false;
return _isSymmetricTree(root1->left,root2->right)
&&_isSymmetricTree(root1->right,root2->left);
}
bool isSymmetric(struct TreeNode* root)
{
if(root==NULL)
return true;
return _isSymmetricTree(root->left,root->right);
}4.二叉树的前序遍历
题目:
力扣https://leetcode.cn/problems/binary-tree-preorder-traversal/
思路:
- 计算二叉树的节点个数
- 创建新的数组，数组大小为二叉树总结点大小
- 前序遍历根，左，右
代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
//计算二叉树的节点个数
int TreeSize(struct TreeNode*root)
{
return root==NULL?0:TreeSize(root->left)+TreeSize(root->right)+1;
}
//创建新的数组，数组大小为二叉树总结点大小
void _preorderTraversal(struct TreeNode*root,int*a,int*pi)
{
if(root==NULL)
return;
//进行前序遍历根，左，右
printf("%d ",root->val);
a[(*pi)++]=root->val;
_preorderTraversal(root->left,a,pi);
_preorderTraversal(root->right,a,pi);
}
int* preorderTraversal(struct TreeNode* root, int* returnSize)
{
int size=TreeSize(root);
printf("size:%d\n",size);
int*a=malloc(sizeof(int)*size);
int i=0;
_preorderTraversal(root,a,&i);
*returnSize=size;
return a;
}5. 二叉树中序遍历
题目:
力扣https://leetcode.cn/problems/binary-tree-inorder-traversal/
思路:
- 计算二叉树的节点个数
- 创建新的数组，数组大小为二叉树总结点大小
- 中序遍历左，根，右
代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
//计算二叉树的节点个数
int TreeSize(struct TreeNode*root)
{
return root==NULL?0:TreeSize(root->left)+TreeSize(root->right)+1;
}
//创建新的数组，数组大小为二叉树总结点大小
void _inorderTraversal(struct TreeNode*root,int*a,int*pi)
{
if(root==NULL)
return;
//进行中序遍历左，根，右
printf("%d ",root->val);
_inorderTraversal(root->left,a,pi);
a[(*pi)++]=root->val;
_inorderTraversal(root->right,a,pi);
}
int* inorderTraversal(struct TreeNode* root, int* returnSize)
{
int size=TreeSize(root);
printf("size:%d\n",size);
int*a=malloc(sizeof(int)*size);
int i=0;
_inorderTraversal(root,a,&i);
*returnSize=size;
return a;
}
6.二叉树后序遍历
题目:
力扣https://leetcode-cn.com/problems/binary-tree-postorder-traversal/
思路:
- 计算二叉树的节点个数
- 创建新的数组，数组大小为二叉树总结点大小
- 后序遍历左，右，根，
代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
//计算二叉树的节点个数
int TreeSize(struct TreeNode*root)
{
return root==NULL?0:TreeSize(root->left)+TreeSize(root->right)+1;
}
//创建新的数组，数组大小为二叉树总结点大小
void _postorderTraversal(struct TreeNode*root,int*a,int*pi)
{
if(root==NULL)
return;
//进行中序遍历左，根，右
printf("%d ",root->val);
_postorderTraversal(root->left,a,pi);
_postorderTraversal(root->right,a,pi);
a[(*pi)++]=root->val;
}
int* postorderTraversal(struct TreeNode* root, int* returnSize)
{
int size=TreeSize(root);
printf("size:%d\n",size);
int*a=malloc(sizeof(int)*size);
int i=0;
_postorderTraversal(root,a,&i);
*returnSize=size;
return a;
}
7.另一颗树的子树
题目:
力扣https://leetcode-cn.com/problems/subtree-of-another-tree/
思路:
思路: 相同的树+递归左子树+递归右子树
代码:
bool isSameTree(struct TreeNode* p, struct TreeNode* q)
{
//根,都为空
if(p==NULL&&q==NULL)
return true;
//其中一个根为空
if(p==NULL||q==NULL)
return false;
//都不为空
if(p->val!=q->val)
return false;
return isSameTree(p->left,q->left)&&isSameTree(p->right,q->right);
}
bool isSubtree(struct TreeNode* root, struct TreeNode* subRoot)
{
if(root==NULL)
return false;
if(isSameTree(root,subRoot))
return true;
return isSubtree(root->left,subRoot)
||isSubtree(root->right,subRoot);
}
8.二叉树遍历
题目:
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思路:
1.建立字符型数组
2.构建一个二叉树，把数组的值输按前序遍历输入到二叉树中，当遇到#时，代表该节点为空
3.将前序遍历好的二叉树，进行中序遍历
代码:
#include
#include
#include
typedef struct TreeNode
{
char val;
struct TreeNode* left;
struct TreeNode* right;
}TreeNode;
TreeNode* CreateTree(char* str,int* pi)
{
if(str[*pi] == '#')
{
(*pi)++;
return NULL;
}
TreeNode* root = (TreeNode*)malloc(sizeof(TreeNode));
root->val = str[(*pi)++];
root->left = CreateTree(str, pi);
root->right = CreateTree(str, pi);
return root;
}
void InOrder(TreeNode* root)
{
if (root == NULL)
{
return;
}
InOrder(root->left);
printf("%c ", root->val);
InOrder(root->right);
}
int main()
{
char str[100];
while(scanf("%s",str)!=EOF)
{
int i = 0;
TreeNode* root = CreateTree(str, &i);
InOrder(root);
}
return 0;
}