二叉树路径求和

二叉树路径和

给定一棵二叉树，二叉树权值为0-9，得出所有根节点到叶节点的路径和。如下图的二叉树，路径和n=137+15=152

     1  

        /\  

3       5

  \

    7

知道用递归遍历i每条路径：但是具体做法还是请教了我对象，下面是我的代码，还有个句话要说就是这道题和《剑指offer》中的找到路径和为22的路径相似，都是要找出所有路径，这道题见http://blog.csdn.net/luxiaoxun/article/details/7537605

下面是求每条路径的路径和：

package binayTree;/** * Created by Administrator on 2015/10/30 0030. */import java.util.ArrayList;import java.util.Stack;/** * Created by chen on 2015/10/30. */class Solution1 {    ArrayList<String> list1 = new ArrayList<String>();    Integer sum = 0;    Stack<Integer> stack=new Stack<Integer>();    public static void main(String[] args) {        Solution1 s = new Solution1();        Stack<Integer> stack=new Stack<Integer>();        Node root = new Node(1);        Node node1 = new Node(3);        Node node2 = new Node(5);        Node node3 = new Node(7);        root.left = node1;        root.right = node2;        node1.right = node3;        System.out.println(s.sumNumbers(root));    }    public int sumNumbers(Node root) {        if (root == null)            return 0;        stack.push(root.value);        // root to this leaf load sum        if (root.left == null && root.right == null) {            int now = 0;            for (Integer i : stack) {                now = now * 10 + i;            }            sum += now;        } else {            if (root.left != null) {                sumNumbers(root.left);                stack.pop();            }            if (root.right != null) {                sumNumbers(root.right);                stack.pop();            }        }        return sum;    }}