Binary Tree Paths

Question:

Given a binary tree, return all root-to-leaf paths.

Example
Given the following binary tree:

   1
 /   \
2     3
 \
  5
All root-to-leaf paths are:

[
  "1->2->5",
  "1->3"
]

Lesson Learned

Code:

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root the root of the binary tree
     * @return all root-to-leaf paths
     */
    private StringBuilder s;
    private List<String> lists;

    public List<String> binaryTreePaths(TreeNode root) {
        s = new StringBuilder();
        lists = new ArrayList<String>();
        helper(root);
        // donot forget return
        return lists;
    }

    public void helper(TreeNode root) {
        // Write your code here
        if (root == null) {
            return;
        }
        s.append("->");
        s.append(root.val);
        if (root.left == null && root.right == null) {
            String str = s.toString();
            lists.add(str.substring(2, str.length()));
        }
        helper(root.left);
        helper(root.right);
        // delete not remove; delete(index start, index end)
        // 有负数的情况所以不能只判断‘-’，必须判断‘->’
        while (s.charAt(s.length() - 2) != '-' || s.charAt(s.length() - 1) != '>') {
            s.delete(s.length()-1, s.length());
        }
        s.delete(s.length()-2, s.length());
    }
}

Updated:

// you can pass whatever you want to the helper method

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root the root of the binary tree
     * @return all root-to-leaf paths
     */
    private String path;
    private List<String> list;
    public List<String> binaryTreePaths(TreeNode root) {
        // Write your code here

        list = new ArrayList<String>();
        if (root == null) {
            return list;
        }
        path = String.valueOf(root.val);
        binaryTreePath(root, path);
        return list;
    }
    // 让list做返回值特别不好，因为返回值一般是对下一轮有影响的，因此把list当做参数传递进来。
    // 不传进来直接作为全局变量也是可以的。
    // path是随路径变化的，因此作为传递函数
    // list只向里面加东西，所以可以作为全局变量存在
    private void binaryTreePath(TreeNode root, String path) {

        if (root.left == null && root.right == null) {
            list.add(path);
            return;
        }
        if (root.left != null) {
            binaryTreePath(root.left, path + "->" + root.left.val);
        }
        if (root.right != null) {
            binaryTreePath(root.right, path + "->" + root.right.val);
        }
    }
}