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

For example, given the following binary tree:

   1
 /   \
2     3
 \
  5

All root-to-leaf paths are:

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

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

思路：這道題我想多了，我用的非遞迴雙棧模擬實現的，其實根本用不著，遞迴實現就行。

遞迴版：

public List<String> binaryTreePaths(TreeNode root) {
    List<String> answer = new ArrayList<String>();
    if (root != null) searchBT(root, "", answer);
    return answer;
}
private void searchBT(TreeNode root, String path, List<String> answer) {
    if (root.left == null && root.right == null) answer.add(path + root.val);
    if (root.left != null) searchBT(root.left, path + root.val + "->", answer);
    if (root.right != null) searchBT(root.right, path + root.val + "->", answer);
}
 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
	public List<String> binaryTreePaths(TreeNode root) {
		LinkedList<TreeNode> is = new LinkedList<TreeNode>();// 記錄元素
		LinkedList<Character> cs = new LinkedList<Character>();// 記錄 L or R
		List<String> res = new ArrayList<String>();
		TreeNode n = root;
		while (n != null || is.size() > 0) {
			while (n != null) {
				is.addLast(n);
				cs.addLast('L');
				n = n.left;
			}
			TreeNode t = is.getLast();
			char type = cs.getLast();
			if (type == 'L') {
				cs.removeLast();
				cs.add('R');
				n = t.right;
			} else {
				if (t.left == null && t.right == null) {
					StringBuilder sb = new StringBuilder();
					for (int i = 0; i < is.size() - 1; i++) {
						sb.append(is.get(i).val + "->");
					}
					sb.append(is.get(is.size() - 1).val);
					res.add(sb.toString());
				}
				is.removeLast();
				cs.removeLast();
				n = null;
			}
		}
		return res;
	}
}