题解 | #二叉树中和为某一值的路径(二)#

本着能不用全局变量就不用全局变量的原则

void dfs(struct TreeNode* root,int target,int** returnColumnSizes,int* returnSize,int count,int* path,int** res){
    if(root==NULL){
        return;
    }
    path[count++]=root->val;
    target-=root->val;
    if(root->right==NULL&&root->left==NULL&&target==0){
        int* temp=(int*)malloc(sizeof(int)*count);
        for(int i=0;i<count;i++){
            temp[i]=path[i];
        }
        res[(*returnSize)]=temp;
        (*returnColumnSizes)[(*returnSize)++]=count;
    }
    dfs(root->left,target,returnColumnSizes,returnSize,count,path,res);
    dfs(root->right,target,returnColumnSizes,returnSize,count,path,res);
    count--;
}


int** FindPath(struct TreeNode* root, int target, int* returnSize, int** returnColumnSizes){
    int** res=(int**)malloc(sizeof(int*)*5000);
    *returnColumnSizes=(int*)malloc(sizeof(int)*5000);
    int* path=(int*)malloc(sizeof(int)*5000);
    int count=0;
    *returnSize=0;
    dfs(root,target,returnColumnSizes,returnSize,count,path,res);
    return res;
}