给定一个二叉树，返回所有从根节点到叶子节点的路径。

说明: 叶子节点是指没有子节点的节点。

示例:

输入:

1
/
2 3

5

输出: [“1->2->5”, “1->3”]

解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> paths = new ArrayList<String>();
        BuildPaths(root, "", paths);
        return paths;
    }

    public void BuildPaths(TreeNode t, String path, List<String> paths){    //paths是一个List集合，所以不能用ArrayList
        if(t == null){
            return;
        }
        StringBuffer sb = new StringBuffer(path);  //暂时存在sb里
        sb.append(Integer.toString(t.val));  //先序遍历
        if(t.left == null && t.right == null){  //遇到叶节点的时候把sb里的字符串全传入paths中
            paths.add(sb.toString());
        }else{
            sb.append("->");        //否则，sb加上->，然后往左右子树递归
            BuildPaths(t.left,sb.toString(),paths); //注意中间的字符串应该是sb中的字符串
            BuildPaths(t.right,sb.toString(),paths);
        }
    }
}