二叉树练习题

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文章目录

一、965. 单值二叉树

1、题目描述


题目链接

2、题目代码

写法一：

/**
 * 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);

写法二：

bool _isUnivalTree(struct TreeNode* root, int val)
    if(root == NULL)
        return true;
    if(root->val != val)
        return false;
    return _isUnivalTree(root->left, val)
            && _isUnivalTree(root->right, val);

 
bool isUnivalTree(struct TreeNode* root)
    if(root == NULL)
        return true;
    int val = root->val;
    return _isUnivalTree(root, val);

二、144. 二叉树的前序遍历

1、题目描述



题目链接
测试用例解析

2、题目代码

/**
 * 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().
 */
int pTSize(struct TreeNode* root)

    return (root==NULL) ? 0 : pTSize(root->left) + pTSize(root->right) + 1;

void _preorderTraversal(struct TreeNode* root,int*a ,int* pi )

    if(root == NULL)
    return;
    a[(*pi)++] = root->val;
    _preorderTraversal(root->left , a, pi);
    _preorderTraversal(root->right , a, pi);

int* preorderTraversal(struct TreeNode* root, int* returnSize)
    int size = pTSize(root);
    int* a = (int*)malloc(sizeof(int) * size);
    *returnSize = size;
    int i=0;
    _preorderTraversal(root , a ,&i );
    return a;

三、104. 二叉树的最大深度

1、题目描述


题目来源

2、题目代码

/**
 * Definition for a binary tree node.
 * struct TreeNode 
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * ;
 */


int maxDepth(struct TreeNode* root)
    if(root == NULL)
    return 0;
    int lenghleft = maxDepth(root->left) + 1;
    int lenghright = maxDepth(root->right) + 1;
    return (lenghleft > lenghright) ? lenghleft : lenghright;

四、226. 翻转二叉树

1、题目描述



题目链接

2、题目代码

/**
 * Definition for a binary tree node.
 * struct TreeNode 
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * ;
 */
void swap(struct TreeNode** px,struct TreeNode** py)

    struct TreeNode* tmp = *px;
    *px = *py;
    *py = tmp;


struct TreeNode* invertTree(struct TreeNode* root)
    if(root == NULL)
    return root;
    swap(&root->left,&root->right);
    invertTree(root->left);
    invertTree(root->right);
    return root;

3、方便理解的图

五、94. 二叉树的中序遍历

1、题目描述



题目链接

2、题目代码

/**
 * 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().
 */
int iTSize(struct TreeNode* root)

    return (root == NULL) ? 0 : iTSize(root->left) + iTSize(root->right) + 1;

void _inorderTraversal(struct TreeNode* root,int* a,int* pi)

    if(root == NULL)return;
    _inorderTraversal(root->left,a,pi);
    a[(*pi)++] = root->val;
    _inorderTraversal(root->right,a,pi);

int* inorderTraversal(struct TreeNode* root, int* returnSize)
    int size = iTSize(root);
    int* a = (int*)malloc(sizeof(int) * size);
    *returnSize = size;
    int i = 0;
    _inorderTraversal( root, a, &i);
    return a;

六、145. 二叉树的后序遍历

1、题目描述



题目链接

2、题目代码

/**
 * 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().
 */
int pTSize(struct TreeNode* root)

    return (root == NULL) ? 0 : pTSize(root->left) + pTSize(root->right) + 1;

void _postorderTraversal(struct TreeNode* root,int* a,int* pi)

    if(root == NULL)return;
    _postorderTraversal(root->left,a,pi);
    _postorderTraversal(root->right,a,pi);
    a[(*pi)++] = root->val;

int* postorderTraversal(struct TreeNode* root, int* returnSize)
    int size = pTSize(root);
    int* a = (int*)malloc(sizeof(int) * size);
    *returnSize = size;
    int i = 0;
    _postorderTraversal( root, a, &i);
    return a;

七、100. 相同的树

1、题目描述



题目链接

2、题目代码

/**
 * Definition for a binary tree node.
 * struct TreeNode 
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * ;
 */


bool isSameTree(struct TreeNode* p, struct TreeNode* q)
    if(p == NULL && q == NULL)
    return true;
    if(p == NULL || q == NULL)
    return false;
    if(p->val != q->val)
    return false;
    return isSameTree(p->left,q->left) && isSameTree(p->right,q->right);

八、KY11 二叉树遍历

1、题目描述


题目链接

2、题目代码

#include<stdio.h>
#include<stdlib.h>
typedef char BTDataType;
typedef struct BinaryTreeNode
    BTDataType _data;
	struct BinaryTreeNode* _left;
	struct BinaryTreeNode* _right;
BTNode;
BTNode* createBT(char* a, int* pi)

    if(a[*pi] == '#')
    
       (*pi)++;
        return NULL;
    
    BTNode* root = (BTNode*)malloc(sizeof(BTNode));
    root->_data = a[*pi];
    (*pi)++;
    root->_left = createBT( a,  pi);
    root->_right = createBT( a,  pi);
    return root;

void inOrderBT(BTNode* root)

    if(root == NULL)
    
        return;
    
    inOrderBT(root->_left);
    printf("%c ",root->_data);
    inOrderBT(root->_right);

int main()

    char a[100];
    while(scanf("%s",a) != EOF)
    
        int i = 0;
        BTNode* root = createBT(a , &i);
        inOrderBT(root);    
        
    
    return 0;

九、110. 平衡二叉树

1、题目描述



题目链接

2、题目代码

写法一：

/**
 * Definition for a binary tree node.
 * struct TreeNode 
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * ;
 */
int BinaryTreeDepth(struct TreeNode* root)

	if (root == NULL)
	
		return 0;
	
	int leftDepth = BinaryTreeDepth(root->left);
	int rightDepth = BinaryTreeDepth(root->right);

	return leftDepth > rightDepth ? leftDepth + 1 : rightDepth + 1;


bool isBalanced(struct TreeNode* root)
    if(root == NULL)
    return true;
    int leftDepth = BinaryTreeDepth(root->left);
	int rightDepth = BinaryTreeDepth(root->right);
    if(abs(leftDepth-rightDepth)>1)
    return false;
    return isBalanced(root->left) && isBalanced(root->right);

写法二：

/**
 * Definition for a binary tree node.
 * struct TreeNode 
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * ;
 */
int maxDepth(struct TreeNode* root)
    return root ? 1 + fmax(maxDepth(root->left) , maxDepth(root->right)) : 0;      

 
bool isBalanced(struct TreeNode* root)
    if(root == NULL)
        return true;
    int left = maxDepth(root->left);
    int right = maxDepth(root->right);
    return abs(left - right) < 2
        && isBalanced(root->left)
        && isBalanced(root->right);

十、572. 另一棵树的子树

1、题目描述




题目链接

2、题目代码

/**
 * Definition for a binary tree node.
 * struct TreeNode 
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * ;
 */
bool isSameTree(struct TreeNode* p, struct TreeNode* q)
    if(p == NULL && q == NULL)
    return true;
    if(p == NULL || q == NULL)
    return false;
    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);

十一、101. 对称二叉树

1、题目描述



题目链接

2、题目代码

/**
 * Definition for a binary tree node.
 * struct TreeNode 
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * ;
 */
bool isSameTree(struct TreeNode* p, struct TreeNode* q)
    if(p == NULL && q == NULL)
    return true;
    if(p == NULL || q == NULL)
    return false;
    if(p->val != q->val)
    return false;
    return isSameTree(p->left,q->right) && isSameTree(p->right,q->left);

bool isSymmetric(struct TreeNode* root)
    if(root == NULL)
    return true;
    return isSameTree(root->left,root->right);

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