题解 | #二叉树中和为某一值的路径(三)#

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return int整型
     */
    int num = 0;
    public int FindPath (TreeNode root, int sum) {
        if(root == null ){
            return num;
        }
        // write code here
        dfs(root,sum);
        FindPath(root.left,sum);
        FindPath(root.right,sum);
        return num;
    }
    public void dfs(TreeNode root,int target){
        if(root == null){
            return;
        }
        target = target - root.val;
        if(target == 0){
            num++;
        }
        dfs(root.left,target);
        dfs(root.right,target);
    }
}