【C语言入门】第十五天

【例题1】938. 二叉搜索树的范围和 - 力扣（LeetCode）

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

int rangeSumBST(struct TreeNode* root, int low, int high){
    if(!root) return 0;
    if(root->val >= low && root->val <= high) 
        return rangeSumBST(root->left,low,high) + rangeSumBST(root->right,low,high) + root->val; 
    else 
        return rangeSumBST(root->left,low,high) + rangeSumBST(root->right,low,high);
}

【例题2】LCR 175. 计算二叉树的深度 - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
int calculateDepth(struct TreeNode* root) {
    if(!root) return 0;
    int count1 = calculateDepth(root->left);
    int count2 = calculateDepth(root->right);
    return 1+ (count1 > count2?count1:count2);
}

【例题3】104. 二叉树的最大深度 - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
int maxDepth(struct TreeNode* root){
    if(!root) return 0;
    int count1 = maxDepth(root->left);
    int count2 = maxDepth(root->right);
    return 1+ (count1 > count2?count1:count2);
}

【例题4】226. 翻转二叉树 - 力扣（LeetCode）

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

struct TreeNode* invertTree(struct TreeNode* root){
   if(!root) return root;
    struct TreeNode* temp = root->left;
    root->left = root->right;
    root->right = temp;
    invertTree(root->left);
    invertTree(root->right);
    return root;
}

【例题5】797. 所有可能的路径 - 力扣（LeetCode）

int** p;
int temp[16];
int len;
//深度优先搜索
void dfs(int cur ,int dest,int** graph, int* graphColSize, int* returnSize, int** returnColumnSizes){
    if(cur == dest){
        int* change = (int*)malloc(sizeof(int)*len);  
        memcpy(change,temp,sizeof(int)*len);
        p[*returnSize] = change;
        (*returnColumnSizes)[(*returnSize)++] = len;
        return;
    }
    for(int i=0;i<graphColSize[cur];i++){
        int xx = graph[cur][i];
        temp[len++] = xx;
        dfs(xx,dest,graph,graphColSize,returnSize,returnColumnSizes);
        len--;
    }
}
int** allPathsSourceTarget(int** graph, int graphSize, int* graphColSize, int* returnSize, int** returnColumnSizes){
    //规定初始状态
    len = 0;
    temp[len++] = 0;
    p = (int**)malloc(sizeof(int*)*7000);
    *returnSize = 0;
    *returnColumnSizes = (int*)malloc(sizeof(int)*7000);
    //实现函数 - 深度优先搜索
    dfs(0, graphSize-1,graph,graphColSize,returnSize,returnColumnSizes);
    return p;
}