import java.util.*;

class TreeNode {
    public char val;
    public TreeNode left;//左孩子的引用
    public TreeNode right;//右孩子的引用
    public TreeNode(char val) {
        this.val = val;
    }
}

public class BinaryTree {
    public static int i = 0;

    public TreeNode createTree(String str) {
        TreeNode root = null;
        if (str.charAt(i) != '#') {
            root = new TreeNode(str.charAt(i));
            i++;
            root.left = createTree(str);
            root.right = createTree(str);
        } else {
            i++;
        }
        return root;
    }

    public TreeNode createTree1() {
        TreeNode A = new TreeNode('A');
        TreeNode B = new TreeNode('B');
        TreeNode C = new TreeNode('C');
        TreeNode D = new TreeNode('D');
        TreeNode E = new TreeNode('E');
        TreeNode F = new TreeNode('F');
        TreeNode G = new TreeNode('G');
        TreeNode H = new TreeNode('H');
        A.left = B;
        A.right = C;
        B.left = D;
        B.right = E;
        C.left = F;
        C.right = G;
        E.right = H;
        return A;
    }

    /*public static int i = 0 ;
    public static TreeNode createTree(String str) {
        TreeNode root = null;
        if(str.charAt(i) != '#') {
            root = new TreeNode(str.charAt(i));
            i++;
            root.left = createTree(str);
            root.right = createTree(str);
        }else {
            //遇到# 就是空树
            i++;
        }
        return root;
    }

    public static void inorder(TreeNode root) {
        if(root == null) {
            return;
        }
        inorder(root.left);
        System.out.print(root.val+" ");
        inorder(root.right);
    }


    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        // 注意 hasNext 和 hasNextLine 的区别
        while (in.hasNextLine()) { // 注意 while 处理多个 case
            String str = in.nextLine();
            TreeNode root = createTree(str);
            inorder(root);
        }
    }*/
    // 前序遍历
    void preOrder(TreeNode root) {
        if (root == null) {
            return;
        }
        System.out.print(root.val + " ");
        preOrder(root.left);
        preOrder(root.right);
    }
    //shift+f6->回车才会生效

    public List<Character> preorderTraversal(TreeNode root) {
        List<Character> retlist = new ArrayList<>();
        if (root == null) {
            return retlist;
        }
        retlist.add(root.val);
        List<Character> leftTree = preorderTraversal(root.left);
        retlist.addAll(leftTree);

        List<Character> rightTree = preorderTraversal(root.right);
        retlist.addAll(rightTree);

        return retlist;
    }

    // 中序遍历
    void inOrder(TreeNode root) {
        if (root == null) {
            return;
        }
        inOrder(root.left);
        System.out.print(root.val + " ");
        inOrder(root.right);
    }

    /*public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> retlist = new ArrayList<>();
        if(root == null) {
            return retlist;
        }
        List<Integer> leftTree = inorderTraversal(root.left);
        retlist.addAll(leftTree);

        retlist.add(root.val);


        List<Integer> rightTree =inorderTraversal(root.right);
        retlist.addAll(rightTree);

        return retlist;
    }*/

    // 后序遍历
    void postOrder(TreeNode root) {
        if (root == null) {
            return;
        }
        postOrder(root.left);
        postOrder(root.right);
        System.out.print(root.val + " ");
    }

    //层序遍历
    public void levelOrder(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        if (root == null) {
            return;
        }
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            System.out.print(cur.val + " ");
            if (cur.left != null) {
                queue.offer(cur.left);
            }
            if (cur.right != null) {
                queue.offer(cur.right);
            }
        }
    }

    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> ret = new ArrayList<>();
        if (root == null) {
            return ret;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> list = new ArrayList<>();
            while (size != 0) {
                TreeNode cur = queue.poll();
                list.add(cur.val);
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
                size--;
            }
            ret.add(list);
        }
        return ret;
    }


    /*public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> retlist = new ArrayList<>();
        if(root == null) {
            return retlist;
        }
        List<Integer> leftTree = postorderTraversal
        (root.left);
        retlist.addAll(leftTree);
        List<Integer> rightTree =postorderTraversal(root.right);
        retlist.addAll(rightTree);
        retlist.add(root.val);
        return retlist;
    }*/

    /**
     * 获取树中节点的个数
     */
    int count = 0;

    int size(TreeNode root) {
        if (root == null) {
            return 0;
        }
        count++;
        size(root.left);
        size(root.right);
        return count;
    }

    // 子问题思路
    int size1(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return 1 + size1(root.left) + size1(root.right);
    }

    // 获取叶子节点的个数
    int countLeaf = 0;

    int getLeafNodeCount(TreeNode root) {
        if (root == null) {
            return 0;
        }
        if (root.left == null && root.right == null) {
            countLeaf++;
        }
        getLeafNodeCount(root.left);
        getLeafNodeCount(root.right);
        return countLeaf;
    }

    //子问题思路

    int getLeafNodeCount1(TreeNode root) {
        if (root == null) {
            return 0;
        }
        if (root.left == null && root.right == null) {
            return 1;
        }
        return getLeafNodeCount1(root.left) + getLeafNodeCount1(root.right);
    }

    // 获取第K层节点的个数
    int getKLevelNodeCount(TreeNode root, int k) {
        if (root == null || k < 1) {
            return 0;
        }
        if (k == 1) {
            return 1;
        }
        return getKLevelNodeCount(root.left, k - 1) + getKLevelNodeCount(root.right, k - 1);
    }

    // 获取二叉树的高度
    int getHeight(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return Math.max(getHeight(root.left), getHeight(root.right)) + 1;

    }

    // 检测值为value的元素是否存在
    TreeNode findVal(TreeNode root, char val) {
        if (root == null) return null;
        if (root.val == val) return root;

        TreeNode ret = findVal(root.left, val);
        if (ret != null) {
            return ret;
        }
        ret = findVal(root.right, val);
        if (ret != null) {
            return ret;
        }
        return null;
    }

    /**
     * 是否为完全二叉树
     *
     * @param root 根节点
     * @return
     */

    boolean isCompleteTree(TreeNode root) {
        if (root == null) return true;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            if (cur != null) {
                queue.offer(cur.left);
                queue.offer(cur.right);
            } else {
                break;
            }
        }
        while (!queue.isEmpty()) {
            TreeNode top = queue.peek();
            if (top != null) {
                return false;
            }
            queue.poll();
        }
        return true;
    }

    public boolean isCompleteTreeOJ(TreeNode root) {
        if (root == null) {
            return true;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            if (cur == null) {
                break;
            }
            if (cur.left != null) {
                queue.offer(cur.left);
            }
            if (cur.right != null) {
                queue.offer(cur.right);
            }
        }
        while (!queue.isEmpty()) {
            TreeNode ret = queue.poll();
            if (ret != null) {
                return false;
            }
        }
        return true;
    }

    TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if (root1 == null) {
            return root2;
        }
        if (root2 == null) {
            return root1;
        }
        //  root1.val=root1.val+root2.val;

        root1.left = mergeTrees(root1.left, root2.left);
        root1.right = mergeTrees(root1.right, root2.right);

        return root1;


    }

    private boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null && q != null || p != null && q == null) {
            return false;
        }
        if (p == null && q == null) {
            return true;
        }
        if (p.val != q.val) {
            return false;
        }
        return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);

    }

    public boolean isSubtree(TreeNode root, TreeNode subRoot) {
        if (root == null || subRoot == null) {
            return false;
        }
        if (isSameTree(root, subRoot)) {
            return true;
        }
        if (isSameTree(root.left, subRoot)) {
            return true;
        }
        if (isSameTree(root.right, subRoot)) {
            return true;
        }
        return false;
    }

    public boolean isSymmetricChild(TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        }
        if (left == null || right == null || left.val != right.val) {
            return false;
        }
        return isSymmetricChild(left.left, right.right) && isSymmetricChild(left.right, right.left);
    }

    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        return isSymmetricChild(root.left, root.right);


    }

    // 解法1：根据二叉搜索树的思路
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        //LCA问题
        if (root == null) {
            return root;
        }
        if (root == p || root == q) {
            return root;
        }
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);

        if (left != null && right != null) {
            return root;
        } else if (left != null) {
            return left;
        } else if (right != null) {
            return right;
        }
        return null;
    }
    // 解法二 根据栈 路径思路

    /**
     * 把root到node的路径放到栈里面
     *
     * @param root  根节点
     * @param node  需要找的节点
     * @param stack 存放路径的栈
     * @return
     */
    public boolean getPath(TreeNode root, TreeNode node, Stack<TreeNode> stack) {
        if (root == null || node == null) {
            return false;
        }
        stack.push(root);
        if (root == node) {
            return true;
        }
        boolean flg = getPath(root.left, node, stack);
        if (flg == true) {
            return true;
        }
        flg = getPath(root.right, node, stack);
        if (flg == true) {
            return true;
        }
        stack.pop();
        return false;
    }

    public TreeNode lowestCommonAncestor1(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return null;
        }
        Stack<TreeNode> stack1 = new Stack<>();
        Stack<TreeNode> stack2 = new Stack<>();
        getPath(root, p, stack1);
        getPath(root, q, stack2);
        int size1 = stack1.size();
        int size2 = stack2.size();
        if (size1 > size2) {
            int size = size1 - size2;
            while (size != 0) {
                stack1.pop();
                size--;
            }
            while (!stack1.empty() && !stack2.empty()) {
                if (stack1.pop() == stack2.pop()) {
                    return stack1.pop();
                } else {
                    stack1.pop();
                    stack2.pop();
                }
            }
        }
        if (size2 > size1) {
            int size = size2 - size1;
            while (size != 0) {
                stack2.pop();
                size--;
            }
            while (!stack1.empty() && !stack2.empty()) {
                if (stack1.pop() == stack2.pop()) {
                    return stack1.pop();
                } else {
                    stack1.pop();
                    stack2.pop();
                }
            }
        }
        return null;

    }
    // 二叉搜素树转换为排好序的双向链表

    /*TreeNode prev=null;
     public void inorder(TreeNode pCur){
         if (pCur==null){
             return;
         }
         inorder(pCur.left);
         pCur.left=prev;
         if (prev!=null){
             prev.right=pCur;
         }
         prev=pCur;
         inorder(pCur.right);
     }
     public TreeNode Convert(TreeNode pRootOfTree) {
         if (pRootOfTree==null){
             return null ;
         }
         inorder(pRootOfTree);
         while (pRootOfTree.left!=null){
             pRootOfTree=pRootOfTree.left;
         }
         return pRootOfTree;

     }*/
    TreeNode prev = null;

    public void inorderOJ(TreeNode root) {
        if (root == null) {
            return;
        }
        inorderOJ(root.left);
        root.left = prev;
        if (prev != null) {
            prev.right = root;
        }
        prev = root;

        inorderOJ(root.right);

    }

    public TreeNode Convert(TreeNode root) {
        if (root == null) {
            return null;
        }
        while (root.left != null) {
            root = root.left;
        }
        return root;

    }

    //根据前序遍历和中序遍历来创建二叉树
    public int preIndex = 0;

    public TreeNode createTreeByPandI(int[] preorder, int[] inorder, int inbegin, int inend) {
        if (inbegin > inend) {
            return null;
        }
        TreeNode root = new TreeNode(preorder[preIndex]);
        int rootIndex = findIndexOfI(inorder, inbegin, inend, preorder[preIndex]);
        if (rootIndex == -1) {
            return null;
        }
        preIndex++;

        root.left = createTreeByPandI(preorder, inorder, inbegin, rootIndex - 1);
        root.right = createTreeByPandI(preorder, inorder, rootIndex + 1, inend);
        return root;
    }

    private int findIndexOfI(int[] inoder, int inbegin, int inend, int key) {
        for (int j = inbegin; j <= inend; j++) {
            if (inoder[j] == key) {
                return j;
            }
        }
        return -1;
    }

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder == null || inorder == null) {
            return null;
        }
        return createTreeByPandI(preorder, inorder, 0, inorder.length - 1);
    }

    //根据后序遍历和中序遍历来创建二叉树
    public int postIndex = 0;

    public TreeNode createTreeByPandI(int[] inorder, int[] postorder, int inbegin, int inend) {
        if (inbegin > inend) {
            return null;
        }
        TreeNode root = new TreeNode(postorder[postIndex]);
        int rootIndex = findIndexOfI(inorder, inbegin, inend, postorder[postIndex]);
        if (rootIndex == -1) {
            return null;
        }
        preIndex--;

        root.right = createTreeByPandI(inorder, postorder, rootIndex + 1, inend);
        root.left = createTreeByPandI(inorder, postorder, inbegin, rootIndex - 1);

        return root;
    }

    private int findIndexOfI(int[] inoder, int inbegin, int inend, int key) {
        for (int j = inbegin; j <= inend; j++) {
            if (inoder[j] == key) {
                return j;
            }
        }
        return -1;
    }

    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (postorder == null || inorder == null) {
            return null;
        }
        postIndex = postorder.length - 1;
        return createTreeByPandI(inorder, postorder, 0, inorder.length - 1);
    }

    // 二叉树创建字符串.
    public void trerToString(TreeNode root, StringBuilder sb) {
        if (root == null) {
            return;
        }
        sb.append(root.val);
        if (root.left != null) {
            sb.append("(");
            trerToString(root.left, sb);
            sb.append(")");
        } else {
            if (root.right == null) {
                return;
            } else {
                sb.append("()");
            }

        }
        if (root.right == null) {
            return;
        } else {
            sb.append("(");
            trerToString(root.right, sb);
            sb.append(")");
        }

    }

    public String tree2str(TreeNode root) {
        if (root == null) {
            return null;
        }
        StringBuilder sb = new StringBuilder();
        trerToString(root, sb);
        return sb.toString();
    }

    public TreeNode increasingBST(TreeNode root) {
        if (root == null) {
            return null;
        }
        List<Integer> list = new ArrayList<>();
        inorderBST(root, list);
        TreeNode prevNode = new TreeNode(-1);
        TreeNode rightNode = prevNode;
        for (int val : list) {
            rightNode.right = new TreeNode(val);
            rightNode = rightNode.right;
        }
        return prevNode.right;

    }

    public void inorderBST(TreeNode root, List<Integer> list) {
        if (root == null) {
            return;
        }
        inorderBST(root.left, list);
        list.add(root.val);
        inorderBST(root.right, list);
    }

    //前序遍历非递归实现
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> ret = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;

        while (cur != null || !stack.empty()) {
            while (cur != null) {
                stack.push(cur);
                ret.add(cur.val);
                cur = cur.left;
            }
            TreeNode top = stack.pop();
            cur = top.right;
        }
        return ret;
    }


    //中序遍历非递归实现
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> ret = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;

        while (cur != null || !stack.empty()) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            TreeNode top = stack.pop();
            ret.add(top.val);
            cur = top.right;
        }
        return ret;
    }

    //后续遍历非递归实现
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> ret = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        TreeNode prev=null;
        while (cur != null || !stack.empty()) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }

            TreeNode top = stack.peek();
            if (top.right == null || top.right==prev) { //如果当前节点的右子树被打印过了，直接就可以弹出了
                stack.pop();
                ret.add(top.val);
                prev=top; //记录一下 最后一次打印的节点
            } else {
                cur = top.right;
            }
        }
        return ret;
    }
}