思路 普通的二叉树需要两种遍历结果才能固定二叉树，而对于BST，得到BST的前序遍历，根据BST的性质，第一个元素值为根节点，小于根节点的元素为左子树，大于根节点的元素为右子树。
class Codec { // Encodes a tree to a single string. //BST的前序遍历结果 public String serialize(TreeNode root) { if (root == null) return ""; StringBuilder sb = new StringBuilder(); helper(root, sb); return sb.substring(0, sb.length() - 1); } private void helper(TreeNode root, StringBuilder sb) { if (root == null) return; //拼接当前节点 sb.append(root.val).append(","); helper(root.left, sb); helper(root.right, sb); } // Decodes your encoded data to tree. public TreeNode deserialize(String data) { if (data == null || data.length() == 0) return null; String[] arr = data.split(","); return builder(arr, 0, arr.length - 1); } private TreeNode builder(String[] arr, int lo, int hi) { if (lo > hi) return null; TreeNode root = new TreeNode(Integer.valueOf(arr[lo])); //找到第一个比首元素大的元素位置 int index = hi + 1; for (int i = lo + 1; i <= hi; i++) { if (Integer.valueOf(arr[i]) > root.val) { index = i; break; } } //递归构建子树 root.left = builder(arr, lo + 1, index - 1); root.right = builder(arr, index, hi); return root; } }