437. Path Sum III 二叉树路径的和3

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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
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){
        int res = 0;
        if(root == null)
            return res;
        if(sum == root.val)
            res++;
        res+=dfs(root.left,sum - root.val);
        res+=dfs(root.right,sum - root.val);
        return res;
    }
}

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