剑指 Offer 34. 二叉树中和为某一值的路径-dfs法

剑指 Offer 34. 二叉树中和为某一值的路径

给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。

叶子节点 是指没有子节点的节点。

示例 1：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]

示例 2：

输入：root = [1,2,3], targetSum = 5
输出：[]

示例 3：

输入：root = [1,2], targetSum = 0
输出：[]

这题其实比较常规的，做的时候，注意一下，一些细节，总体使用深度优先遍历去解决这题，可以遍历每条路径，并相对保留路径信息：

/** * 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(). */
 int path[1000];
int size;
void dfs(struct TreeNode* root,int target,int path_now_size,int sum,int **re,int** returnColumnSizes){
    
 // printf("sum %d ",sum);
    if(root){
    
    // printf("val %d path_now_size %d",root->val);
        path[path_now_size]=root->val;
       
          sum=sum+root->val;

         if(sum==target&&root->left==NULL&&root->right==NULL){
    
               re[size]=(int *)malloc(sizeof(int)*(path_now_size+1));
              (*returnColumnSizes)[size]=path_now_size+1;
               int i;
             for(i=0;i<path_now_size+1;i++){
    
                // printf("%d ",path[i]);
                 re[size][i]=path[i];

             }
             size++;
           }
        dfs(root->left,target,path_now_size+1,sum,re,returnColumnSizes);
     dfs(root->right,target,path_now_size+1,sum,re,returnColumnSizes);



    }


}
#define maxsize 1000


int** pathSum(struct TreeNode* root, int target, int* returnSize, int** returnColumnSizes){
    
    int**re=(int**)malloc(sizeof(int *)*maxsize);
    *returnColumnSizes=(int*)malloc(sizeof(int )*maxsize);
    size=0;
      dfs(root,target,0,0,re,returnColumnSizes);
      *returnSize=size;
      printf("szie %d ",size);
      return re;



}