Interview Question 04.12. Summation Path-dfs+Auxiliary Array Method

面试题 04.12. 求和路径

给定一棵二叉树,其中每个节点都含有一个整数数值(该值或正或负).设计一个算法,打印节点数值总和等于某个给定值的所有路径的数量.注意,路径不一定非得从二叉树的根节点或叶节点开始或结束,但是其方向必须向下(只能从父节点指向子节点方向).

示例:
给定如下二叉树,以及目标和 sum = 22,

          5
         / \
        4   8
       /   / \
      11  13  4
     /  \    / \
    7    2  5   1

返回:

3
解释：和为 22 的路径有：[5,4,11,2], [5,8,4,5], [4,11,7]

The code for solving this problem is as follows,Still a good topic,It is recommended not to use the brute force solution directly：

/** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */
int count;
 void  dfs(struct TreeNode*root,int sum,int val,int *r,int h){
    
     if(root){
    
        
         r[h]=root->val;
         if(root->left){
    
             dfs(root->left,sum,root->val+val,r,h+1);
         }
          if(root->right){
    
             dfs(root->right,sum,root->val+val,r,h+1);
         }
         int eval=root->val+val;
         if(eval==sum){
    
             count++;
         }
       
         for(int i=0;i<h;i++){
    
             eval=eval-r[i];
             if(eval==sum){
    
                 count++;
             }
         }
       



     }
 }


int pathSum(struct TreeNode* root, int sum){
    
    int *r=(int *)malloc(sizeof(int)*1000);
    count=0;
    dfs(root,sum,0,r,0);
    return count;

}