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"]

遇到二叉树的题一般都会用到递归遍历，这题就是一个变形，要找所以的路径，思路就是从root开始一直记录路径上的所有节点，然后当遍历到leaf的时候，我们就把路径加入到result中。
把path存在字符串中做递归刚开始也没想到，后来看了别人的思路觉得代码写起来简洁很多，所有就you'hua'le'xia

/**
 * 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
     */
    public List<String> binaryTreePaths(TreeNode root) {
        // Write your code here
        List<String> result = new ArrayList<String>();
        if (root != null) {
            findPath(result, root, String.valueOf(root.val));
        }
        
        return result;
    }
    
    public void findPath(List<String> result, TreeNode p, String path) {
        if(p == null) {
            return ;
        }
        if (p.left == null && p.right == null) {
            result.add(path);  
            return ;
        }
        
        if(p.left != null) {
            findPath(result, p.left, path + "->" + String.valueOf(p.left.val));
        }
        
        if(p.right != null) {
            findPath(result, p.right, path + "->" + String.valueOf(p.right.val));
        }
        
    }
}