9.path Sum III（路径和 III）

Level：

Easy

题目描述：

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

思路分析：

由于题目中所说的路径，不局限于从根到叶子节点，任何一个节点都可以作为路径的开始节点和终止节点，所以我们以根节点作为开始节点，查找和为sum的路径数，然后分别以根的左孩子和右孩子为起始节点去查找和为sum的路径数，依次递归向下推导，得到最终的结果。

代码：

/**public class TreeNode{

    int vla;

    TreeNode left;

    TreeNode right;

    public TreeNode(int val){

        this.val=val;

    }

}*/

public class Soulution{

    public int pathSum(TreeNode root,int sum){

        if(root==null)

            return 0;

        int res=0;

        res=pathCheck(root,sum);

        res=res+pathSum(root.left,sum);

        res=res+pathSum(root.right,sum);

        return res;

    }

    public int pathCheck(TreeNode root,int sum){

        if(root==null)

            return 0;

        int count=0;

        if(sum==root.val)//当sum等于root.val时证明存在一条路径和为sum

            count++;

        count=count+pathCheck(root.left,sum-root.val);

        count=count+pathCheck(root.right,sum-root.val);

        return count;

    }

}

巴特西

9.path Sum III（路径和 III）

Level：

题目描述：

代码：

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