树——练习题

2023-12-23 20:44:46

1. ?二叉树的前序遍历

代码实现：

/**
 * 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().
 */

#define MAX_NODE 100

// 递归
void pre_order1(struct TreeNode* root, int* res, int* returnSize) {
    if (root == NULL) { // 终止条件
        return ;
    }
    res[(*returnSize)++] = root->val;
    pre_order1(root->left, res, returnSize);
    pre_order1(root->right, res, returnSize);
}

// 迭代（非递归） 深度优先搜索---借用栈
struct TreeNode* track[MAX_NODE];
int top = -1;
void pre_order2(struct TreeNode* root, int* res, int* returnSize) {
    if (root == NULL) {
        return ;
    }
    track[++top] = root;
    while (top != -1) { // 栈不为空则执行while循环
        struct TreeNode* r = track[top]; // 记录栈顶
        res[(*returnSize)++] = r->val;
        top--;
        if (r->right) { // 右子树先进栈
            track[++top] = r->right;
        }
        if (r->left) {
            track[++top] = r->left;
        }
    }
}

int* preorderTraversal(struct TreeNode* root, int* returnSize) {
    int* res = malloc(sizeof(int) * MAX_NODE);
    *returnSize = 0;
    pre_order1(root, res, returnSize);
    pre_order2(root, res, returnSize);
    return res;
}

2. 二叉树的中序遍历

代码实现：

/**
 * 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().
 */

#define MAX_NODE 100

// 递归
void mid_order1(struct TreeNode* root, int* res, int* returnSize) {
    if (root == NULL) { // 终止条件
        return ;
    }
    mid_order1(root->left, res, returnSize);
    res[(*returnSize)++] = root->val;
    mid_order1(root->right, res, returnSize);
}

// 迭代（非递归） 深度优先搜索---借用栈
struct TreeNode* stack[MAX_NODE];
int top = -1;
void mid_order2(struct TreeNode* root, int* res, int* returnSize) {
    if (root == NULL) {
        return ;
    }
    struct TreeNode* p = root;
    struct TreeNode* q = NULL;
    while (p || top != -1) {
		if (p) {
			stack[++top] = p;
			p = p->left;
		} else {
			q = stack[top];
			res[(*returnSize)++] = q->val;
            top--;
			p = q->right;
		}
	}
}

int* inorderTraversal(struct TreeNode* root, int* returnSize) {
    int* res = malloc(sizeof(int) * MAX_NODE);
    *returnSize = 0;
    mid_order1(root, res, returnSize);
    mid_order2(root, res, returnSize);
    return res;
}

3.?二叉树的后序遍历

代码实现：

/**
 * 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().
 */

#define MAX_NODE 100

// 递归
void post_order1(struct TreeNode* root, int* res, int* returnSize) {
    if (root == NULL) { // 终止条件
        return ;
    }
    post_order1(root->left, res, returnSize);
    post_order1(root->right, res, returnSize);
    res[(*returnSize)++] = root->val;
}

// 迭代（非递归） 深度优先搜索---借用栈
/*
第一种方法：
    先序遍历是 中左右 --->调整代码左右循序--->中右左--->反转result数组--->左右中
    后序遍历是 左右中
*/
/*
第二种方法：
    1、沿着根的左孩子，依次入栈，直到左孩子为空;
    2、读栈顶元素进行判定，若右孩子不空且未被访问，将右孩子执行第一步;
    3、栈顶元素出栈。
*/
struct TreeNode* stack[MAX_NODE];
int top = -1; // 栈顶 
void post_order2(struct TreeNode* root, int* res, int* returnSize) {
    struct TreeNode *p = root, *r = NULL; //r标记最近访问过的结点
	while (p || top != -1) {
		if (p) {
			stack[++top] = p;
			p = p->left;
		} else {
			p = stack[top];
			// 获取栈顶元素赋值给p
			// ?case one?
			if (p->right && p->right != r) { //若右孩子存在且未被访问
				p = p->right; // 就让右孩子
				stack[++top] = p; // 入栈
				p = p->left; // 让右孩子向左
				// 上面三句意思就是让右孩子的左孩子一直入栈，一直向左走
			} else {
			    // ?case two?
                res[(*returnSize)++] = stack[top--]->val; // 右孩子为空或未被访问过，就出栈
				r = p;  //r标记最近访问结点
				p = NULL;  //r置空
				// 置空原因：因为这个结点已经出栈了
			}
		}
	}
}

int* postorderTraversal(struct TreeNode* root, int* returnSize) {
    int* res = malloc(sizeof(int) * MAX_NODE);
    *returnSize = 0;
    post_order1(root, res, returnSize);
    post_order2(root, res, returnSize);
    return res;
}

4. 二叉树的层序遍历

代码实现：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
// returnColumnSizes：用于存储每个数组中元素个数
int** levelOrder(struct TreeNode* root, int* returnSize, int** returnColumnSizes) {
    *returnSize = 0;
    if (root == NULL) {
        return NULL;
    }
    int** res = (int**)malloc(sizeof(int*) * 2000);
    *returnColumnSizes = (int*)malloc(sizeof(int) * 2000);
    struct TreeNode* queue[2001]; // 创建一个队列 注意：队头不放元素，长度须+1
    int head = 0; // 队头
    int tail = 0; // 队尾
    queue[++tail] = root; //head:不存放数据，tail:存放数据
    while (head < tail) {
        int colSize = 0, last = tail;
        res[*returnSize] = (int*)malloc(sizeof(int) * (last - head));
        while (head < last) {
            struct TreeNode* node = queue[++head];
            res[*returnSize][colSize++] = node->val;
            if (node->left != NULL) {
                queue[++tail] = node->left;
            }
            if (node->right != NULL) {
                queue[++tail] = node->right;
            }       
        }
        (*returnColumnSizes)[*returnSize] = colSize;
        (*returnSize)++;
    }
    return res;
}

5. 二叉树的层序遍历||

思路：相对于102.二叉树的层序遍历，就是最后把res数组反转一下就可以了

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
// returnColumnSizes：用于存储每个数组中元素个数
int** levelOrderBottom(struct TreeNode* root, int* returnSize, int** returnColumnSizes) {
    *returnSize = 0;
    if (root == NULL) {
        return NULL;
    }
    int** res = (int**)malloc(sizeof(int*) * 2000);
    *returnColumnSizes = (int*)malloc(sizeof(int) * 2000);
    struct TreeNode* queue[2001]; // 创建一个队列 注意：队头不放元素，长度须+1
    int head = 0; // 队头
    int tail = 0; // 队尾
    queue[++tail] = root; //head:不存放数据，tail:存放数据
    while (head < tail) {
        int colSize = 0, last = tail;
        res[*returnSize] = (int*)malloc(sizeof(int) * (last - head));
        while (head < last) {
            struct TreeNode* node = queue[++head];
            res[*returnSize][colSize++] = node->val;
            if (node->left) {
                queue[++tail] = node->left;
            }
            if (node->right) {
                queue[++tail] = node->right;
            }       
        }
        (*returnColumnSizes)[*returnSize] = colSize;
        (*returnSize)++;
    }

    // 反转res数组
    int** ret = (int**)malloc(sizeof(int*) * (*returnSize));
    int nums[*returnSize];
    for (int i = 0; i < *returnSize; i++) {
        nums[i] = (*returnColumnSizes)[i];
    }
    for (int i = 0; i < *returnSize; i++) {
        (*returnColumnSizes)[i] = nums[*returnSize - i - 1];
    }

    for (int i = 0; i < *returnSize; i++) {
        ret[i] = (int*)malloc(sizeof(int) * nums[*returnSize - i - 1]);
        for (int j = 0; j < nums[*returnSize - i - 1]; j++) {
            ret[i][j] = res[*returnSize - i - 1][j];
        }
    }
    free(res);
    return ret;
}

6.?二叉树的右视图

思路：层序遍历的时候，判断是否遍历到单层的最后面的元素，如果是，就放进result数组中，随后返回result就可以了

/**
 * 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* rightSideView(struct TreeNode* root, int* returnSize) {
    int* ret = malloc(sizeof(int) * 100);
    memset(ret, 0, sizeof(int) * 100);
    *returnSize = 0;
    if (root == NULL) {
        return ret;
    }
    struct TreeNode* queue[101]; // 创建一个队列
    int head = 0; // 对头
    int tail = 0; // 队尾
    queue[++tail] = root; //head:不存放数据，tail:存放数据
    while (head < tail) {
        int last = tail;
        while (head < last) {
            struct TreeNode* node = queue[++head];
            if (head == last) {
                ret[(*returnSize)++] = node->val;
            }
            if (node->left) {
                queue[++tail] = node->left;
            }
            if (node->right) {
                queue[++tail] = node->right;
            }
        }
    }
    return ret;
}


/**
 * 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().
 */
double* averageOfLevels(struct TreeNode* root, int* returnSize) {
    double* ret = malloc(sizeof(double) * 10000);
    memset(ret, 0, sizeof(int) * 10000);
    *returnSize = 0;
    if (root == NULL) {
        return ret;
    }
    int queue[10001]; // 创建一个队列
    int head = 0; // 对头
    int tail = 0; // 队尾
    queue[++tail] = root; //head:不存放数据，tail:存放数据
    while (head < tail) {
        int last = tail;
        double sum = 0;
        int size = tail - head; // 每层结点个数
        while (head < last) {
            struct TreeNode* node = queue[++head];
            sum += node->val;
            if (node->left) {
                queue[++tail] = node->left;
            }
            if (node->right) {
                queue[++tail] = node->right;
            }
        }
        ret[(*returnSize)++] = sum / size;
    }
    return ret;
}

7. 二叉树的层平均值

代码实现：

/**
 * 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().
 */
double* averageOfLevels(struct TreeNode* root, int* returnSize) {
    double* ret = malloc(sizeof(double) * 10000);
    memset(ret, 0, sizeof(int) * 10000);
    *returnSize = 0;
    if (root == NULL) {
        return ret;
    }
    struct TreeNode* queue[10001]; // 创建一个队列
    int head = 0; // 对头
    int tail = 0; // 队尾
    queue[++tail] = root; //head:不存放数据，tail:存放数据
    while (head < tail) {
        int last = tail;
        double sum = 0;
        int size = tail - head; // 每层结点个数
        while (head < last) {
            struct TreeNode* node = queue[++head];
            sum += node->val;
            if (node->left) {
                queue[++tail] = node->left;
            }
            if (node->right) {
                queue[++tail] = node->right;
            }
        }
        ret[(*returnSize)++] = sum / size;
    }
    return ret;
}

8. N叉树的前序遍历

9. 在每个树行中找最大值 515

10. 填充每个节点的下一个右侧节点指针 116

11.?填充每个节点的下一个右侧节点指针|| 117

12. 二叉树的最大深度? 104

13. 二叉树的最小深度 111

14. 翻转二叉树

思路：

在翻转时，需要从根节点往下先交换左右子树，然后再递归到左右子树进行相同的操作

? ? ? ? （前序遍历 and 层序遍历）

在翻转时，需要从下往上先交换左右子树，然后再递归到父节点进行交换（后序遍历）
中序遍历不可以（原因：先翻转左子树，然后左右子树交换，这时候再翻转的右子树是之前的左子树）

层序遍历：

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

// 交换左右子树
void swap(struct TreeNode* root) {
    struct TreeNode* l = root->left;
    struct TreeNode* r = root->right;
    root->left = r;
    root->right = l;
}

struct TreeNode* invertTree(struct TreeNode* root) {
    if (root == NULL) {
        return root;
    }
    struct TreeNode* queue[101]; // 创建一个队列
    int head = 0; // 队头
    int tail = 0; // 队尾
    queue[++tail] = root; // head:不存放数据，tail:存放数据
    while (head < tail) {
        struct TreeNode* node = queue[++head];
        swap(node); // 翻转
        if (node->left) {
            queue[++tail] = node->left;
        }
        if (node->right) {
            queue[++tail] = node->right;
        }
    }
    return root;
}

前序遍历：

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

// 交换左右子树
void swap(struct TreeNode* root) {
    struct TreeNode* l = root->left;
    struct TreeNode* r = root->right;
    root->left = r;
    root->right = l;
}

// 递归：先序遍历
void pre_order(struct TreeNode* root) {
    if (root == NULL) {
        return ;
    }
    swap(root); // 交换
    if (root->left) {
        pre_order(root->left);
    }
    if (root->right) {
        pre_order(root->right);
    }
}

struct TreeNode* invertTree(struct TreeNode* root) {
    if (root == NULL) {
        return root;
    }
    pre_order(root);
    return root;
}

后序遍历：

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

// 交换左右子树
void swap(struct TreeNode* root) {
    struct TreeNode* l = root->left;
    struct TreeNode* r = root->right;
    root->left = r;
    root->right = l;
}

// 递归：后序遍历
void pre_order(struct TreeNode* root) {
    if (root == NULL) {
        return ;
    }
    if (root->left) {
        pre_order(root->left);
    }
    if (root->right) {
        pre_order(root->right);
    }
    swap(root); // 交换
}

struct TreeNode* invertTree(struct TreeNode* root) {
    if (root == NULL) {
        return root;
    }
    pre_order(root);
    return root;
}

文章来源:https://blog.csdn.net/qq_68915288/article/details/135162274
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