通过将节点值左移数位将二叉树转换为BST

给定正整数的二叉树。任务是使用节点数字上的左移操作将其转换为BST 。如果无法将二进制树转换为BST，则打印-1 。

例子：

Input:                      443 
                              /        \  
                          132      543 
                                    /      \ 
                                343     237

Output:                   344 
                               /        \ 
                         132       435 
                                   /        \ 
                               433       723

Explanation:   443 → left shift two times → 344 
                         132 → zero shift operations → 132 
                         543 → left shift one time → 435 
                         343 → left shift one time → 433 
                         237 → left shift two times → 723

Input:                    43 
                         /        \ 
                      56         45

Output: -1

方法：因为对BST进行有序遍历总是会生成排序的序列，所以这种想法是对二叉树进行有序遍历。请按照以下步骤解决问题：

• 初始化一个变量，例如prev = -1 ，以跟踪上一个节点的值。
• 然后，使用“有序遍历”遍历树，并尝试通过向左移动其数字来将每个节点转换为大于先前值的最低可能值。
• 执行完这些操作后，检查二进制树是否为BST 。
• 如果发现是真的，则打印树。否则，打印-1 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Structure of a Node
struct TreeNode
{
   int val;
   TreeNode *left, *right;
 
    TreeNode(int key)
    {
        val = key;
        left = NULL;
        right = NULL;
    }
};
 
// Function to check if
// the tree is BST or not
bool isBST(TreeNode *root, TreeNode *l = NULL, TreeNode *r = NULL)
{
 
    // Base condition
    if (!root)
        return true;
 
    // Check for left child
    if (l && root->val <= l->val)
        return false;
 
    // Check for right child
    if (r && root->val >= r->val)
        return false;
 
    // Check for left and right subtrees
    return isBST(root->left, l, root) and isBST(root->right, root, r);
}
 
// Function to convert a tree to BST
// by left shift of its digits
void ConvTree(TreeNode *root,int &prev)
{
 
    // If root is NULL
    if (!root)
        return;
 
    // Recursively modify the
    // left subtree
    ConvTree(root->left,prev);
 
    // Initialise optimal element
    // to the initial element
    int optEle = root->val;
 
    // Convert it into a string
    string strEle = to_string(root->val);
 
    // For checking all the
    // left shift operations
    bool flag = true;
    for (int idx = 0; idx < strEle.size(); idx++)
    {
 
        // Perform left shift
        int shiftedNum = stoi(strEle.substr(idx) + strEle.substr(0, idx));
 
        // If the number after left shift
        // is the minimum and greater than
        // its previous element
 
        // first value >= prev
 
        // used to setup initialize optEle
        // cout<= prev and flag)
        {
            optEle = shiftedNum;
            flag = false;
          }
 
        if (shiftedNum >= prev)
            optEle = min(optEle, shiftedNum);
      }
    root->val = optEle;
    prev = root->val;
   
    // Recursively solve for right
    // subtree
    ConvTree(root->right,prev);
}
 
// Function to print level
// order traversal of a tree
void levelOrder(TreeNode *root)
{
    queue que;
    que.push(root);
    while(true)
    {
        int length = que.size();
        if (length == 0)
            break;
        while (length)
        {
            auto temp = que.front();
            que.pop();
            cout << temp->val << " ";
            if (temp->left)
                que.push(temp->left);
            if (temp->right)
                que.push(temp->right);
            length -= 1;
          }
          cout << endl;
        }
   
        // new line
        cout << endl;
}
 
// Driver Code
int main()
{
 
  TreeNode *root = new TreeNode(443);
  root->left = new TreeNode(132);
  root->right = new TreeNode(543);
  root->right->left = new TreeNode(343);
  root->right->right = new TreeNode(237);
 
  // Converts the tree to BST
  int prev = -1;
  ConvTree(root, prev);
 
  // If tree is converted to a BST
  if (isBST(root, NULL, NULL))
  {
      // Print its level order traversal
      levelOrder(root);
    }
  else
  {
     
      // not possible
      cout << (-1);
    }
}
 
// This code is contributed by mohit kumar 29

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
static int prev;
 
// Structure of a Node
static class TreeNode
{
    int val;
    TreeNode left, right;
     
    TreeNode(int key)
    {
        val = key;
        left = null;
        right = null;
    }
};
 
// Function to check if
// the tree is BST or not
static boolean isBST(TreeNode root, TreeNode l,
                     TreeNode r)
{
     
    // Base condition
    if (root == null)
        return true;
 
    // Check for left child
    if (l != null && root.val <= l.val)
        return false;
 
    // Check for right child
    if (r != null && root.val >= r.val)
        return false;
 
    // Check for left and right subtrees
    return isBST(root.left, l, root) &
           isBST(root.right, root, r);
}
 
// Function to convert a tree to BST
// by left shift of its digits
static void ConvTree(TreeNode root)
{
     
    // If root is null
    if (root == null)
        return;
         
    // Recursively modify the
    // left subtree
    ConvTree(root.left);
 
    // Initialise optimal element
    // to the initial element
    int optEle = root.val;
 
    // Convert it into a String
    String strEle = String.valueOf(root.val);
 
    // For checking all the
    // left shift operations
    boolean flag = true;
     
    for(int idx = 0; idx < strEle.length(); idx++)
    {
         
        // Perform left shift
        int shiftedNum = Integer.parseInt(
            strEle.substring(idx) +
            strEle.substring(0, idx));
 
        // If the number after left shift
        // is the minimum and greater than
        // its previous element
 
        // first value >= prev
 
        // used to setup initialize optEle
        // System.out.print(prev<= prev && flag)
        {
            optEle = shiftedNum;
            flag = false;
        }
         
        if (shiftedNum >= prev)
            optEle = Math.min(optEle, shiftedNum);
    }
    root.val = optEle;
    prev = root.val;
   
    // Recursively solve for right
    // subtree
    ConvTree(root.right);
}
 
// Function to print level
// order traversal of a tree
static void levelOrder(TreeNode root)
{
    Queue que = new LinkedList();
    que.add(root);
     
    while(true)
    {
        int length = que.size();
         
        if (length == 0)
            break;
             
        while (length > 0)
        {
            TreeNode temp = que.peek();
            que.remove();
            System.out.print(temp.val + " ");
             
            if (temp.left != null)
                que.add(temp.left);
            if (temp.right != null)
                que.add(temp.right);
                 
            length -= 1;
        }
        System.out.println();
    }
     
    // New line
    System.out.println();
}
 
// Driver Code
public static void main(String[] args)
{
    TreeNode root = new TreeNode(443);
    root.left = new TreeNode(132);
    root.right = new TreeNode(543);
    root.right.left = new TreeNode(343);
    root.right.right = new TreeNode(237);
     
    // Converts the tree to BST
    prev = -1;
    ConvTree(root);
     
    // If tree is converted to a BST
    if (isBST(root, null, null))
    {
         
        // Print its level order traversal
        levelOrder(root);
    }
    else
    {
         
        // not possible
        System.out.print(-1);
    }
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 program for the above approach
 
# Structure of a Node
class TreeNode:
    def __init__(self, key):
        self.val = key
        self.left = None
        self.right = None
 
# Function to check if
# the tree is BST or not
def isBST(root, l = None, r = None):
 
    # Base condition
    if not root:
        return True
 
    # Check for left child
    if l and root.val <= l.val:
        return False
 
    # Check for right child
    if r and root.val >= r.val:
        return False
 
    # Check for left and right subtrees
    return isBST(root.left, l, root) and isBST(root.right, root, r)
 
# Function to convert a tree to BST
# by left shift of its digits
def ConvTree(root):
    global prev
 
    # If root is NULL
    if not root:
        return
 
    # Recursively modify the
    # left subtree
    ConvTree(root.left)
 
    # Initialise optimal element
    # to the initial element
    optEle = root.val
 
    # Convert it into a string
    strEle = str(root.val)
 
    # For checking all the
    # left shift operations
    flag = True
 
    for idx in range(len(strEle)):
 
        # Perform left shift
        shiftedNum = int(strEle[idx:] + strEle[:idx])
 
        # If the number after left shift
        # is the minimum and greater than
        # its previous element
 
        # first value >= prev
 
        # used to setup initialize optEle
        if shiftedNum >= prev and flag:
            optEle = shiftedNum
            flag = False
 
        if shiftedNum >= prev:
            optEle = min(optEle, shiftedNum)
 
    root.val = optEle
    prev = root.val
    # Recursively solve for right
    # subtree
    ConvTree(root.right)
 
# Function to print level
# order traversal of a tree
def levelOrder(root):
    que = [root]
    while True:
        length = len(que)
        if not length:
            break
 
        while length:
            temp = que.pop(0)
            print(temp.val, end =' ')
            if temp.left:
                que.append(temp.left)
            if temp.right:
                que.append(temp.right)
            length -= 1
        # new line
        print()
 
# Driver Code
 
root = TreeNode(443)
root.left = TreeNode(132)
root.right = TreeNode(543)
root.right.left = TreeNode(343)
root.right.right = TreeNode(237)
 
# Initialize to
# previous element
prev = -1
 
# Converts the tree to BST
ConvTree(root)
 
# If tree is converted to a BST
if isBST(root, None, None):
 
    # Print its level order traversal
    levelOrder(root)
else:
    # not possible
    print(-1)

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
     
static int prev;
 
// Structure of a Node
class TreeNode
{
    public int val;
    public TreeNode left, right; 
    public TreeNode(int key)
    {
        val = key;
        left = null;
        right = null;
    }
};
 
// Function to check if
// the tree is BST or not
static bool isBST(TreeNode root, TreeNode l,
                     TreeNode r)
{
     
    // Base condition
    if (root == null)
        return true;
 
    // Check for left child
    if (l != null && root.val <= l.val)
        return false;
 
    // Check for right child
    if (r != null && root.val >= r.val)
        return false;
 
    // Check for left and right subtrees
    return isBST(root.left, l, root) &
           isBST(root.right, root, r);
}
 
// Function to convert a tree to BST
// by left shift of its digits
static void ConvTree(TreeNode root)
{
     
    // If root is null
    if (root == null)
        return;
         
    // Recursively modify the
    // left subtree
    ConvTree(root.left);
 
    // Initialise optimal element
    // to the initial element
    int optEle = root.val;
 
    // Convert it into a String
    String strEle = String.Join("", root.val);
 
    // For checking all the
    // left shift operations
    bool flag = true;
     
    for(int idx = 0; idx < strEle.Length; idx++)
    {
         
        // Perform left shift
        int shiftedNum = Int32.Parse(
            strEle.Substring(idx) +
            strEle.Substring(0, idx));
 
        // If the number after left shift
        // is the minimum and greater than
        // its previous element
 
        // first value >= prev
 
        // used to setup initialize optEle
        // Console.Write(prev<= prev && flag)
        {
            optEle = shiftedNum;
            flag = false;
        }
         
        if (shiftedNum >= prev)
            optEle = Math.Min(optEle, shiftedNum);
    }
    root.val = optEle;
    prev = root.val;
   
    // Recursively solve for right
    // subtree
    ConvTree(root.right);
}
 
// Function to print level
// order traversal of a tree
static void levelOrder(TreeNode root)
{
    Queue que = new Queue();
    que.Enqueue(root);
     
    while(true)
    {
        int length = que.Count;
        if (length == 0)
            break;     
        while (length > 0)
        {
            TreeNode temp = que.Peek();
            que.Dequeue();
            Console.Write(temp.val + " ");      
            if (temp.left != null)
                que.Enqueue(temp.left);
            if (temp.right != null)
                que.Enqueue(temp.right);       
            length -= 1;
        }
        Console.WriteLine();
    }
     
    // New line
    Console.WriteLine();
}
 
// Driver Code
public static void Main(String[] args)
{
    TreeNode root = new TreeNode(443);
    root.left = new TreeNode(132);
    root.right = new TreeNode(543);
    root.right.left = new TreeNode(343);
    root.right.right = new TreeNode(237);
     
    // Converts the tree to BST
    prev = -1;
    ConvTree(root);
     
    // If tree is converted to a BST
    if (isBST(root, null, null))
    {
         
        // Print its level order traversal
        levelOrder(root);
    }
    else
    {
         
        // not possible
        Console.Write(-1);
    }
}
}
  
// This code is contributed by shikhasingrajput

输出：

344 
132 435 
433 723

时间复杂度： O(N)
辅助空间： O(N)