題目描述

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

解題思路

這道題可以借用之前Path Sum的思想，不過它需要兩層深度優先遍歷。

先從頭結點開始遍歷，求得滿足條件的路徑的個數後，再分別遍歷其左右子樹，求得滿足條件的路徑再累加。

第一種方法：用佇列層次遍歷樹的每個節點，再從每個節點開始dfs求出滿足條件的路徑，將所有的累加就得到結果。

程式實現

public class Solution {


    public int pathSum(TreeNode root, int sum) {
        if(root==null)
            return 0;
        return dfs(root,sum)+pathSum(root.left,sum)+pathSum(root.right,sum);
    }


   private  int dfs(TreeNode root,int
 
 sum){

       if(root==null)
           return 0;
       int count=0;
       if(sum==root.val)
           count++;
       return count+dfs(root.left,sum-root.val)+dfs(root.right,sum-root.val);
   }
}