给定一个二叉树，任务是通过向左旋转每个节点任意次数来修改树，这样每个级别都由从左到右递增的节点值组成。如果无法按升序排列任何级别的节点值，则打印“-1” 。
例子：
Input:
341
/ \
241 123
/ \ \
324 235 161
Output:
341
/ \
124 231
/ \ \
243 352 611
Explanation:
Level 1: The value of the root node is 341, having all digits sorted.
Level 2: The node values are 241 and 123. Left rotating 241 twice and 231 once modifies the node values to 124 and 231.
Level 3: The node values are 342, 235 and 161. Left rotating digits of 342 once, 235 once and 161 once modifies the node values to {243, 352, 611} respectively.
Input:
12
/ \
89 15
Output: -1
方法：给定的问题可以通过执行 Level Order Traversal 并左旋节点值的数字来解决给定的问题，以使每个级别的值按升序排列。请按照以下步骤解决问题：
- 初始化一个队列，比如说Q ，用于执行层序遍历。
- 将树的根节点推入队列。
- 迭代直到队列不为空并执行以下步骤：
- 找到队列的大小并将其存储在变量L 中。
- 初始化一个变量，比如prev ，用于存储树的当前级别中的前一个元素。
- 迭代范围[0, L]并执行以下步骤：
- 弹出队列的前端节点并将其存储在一个变量中，比如temp 。
- 现在，左移元素temp ，如果存在任何大于prev 且更接近prev 的排列，则将当前节点的值更新为temp的值。
- 将prev的值更新为temp的当前值。
- 如果temp有左孩子或右孩子，则将其推入队列。
- 经过上述步骤后，如果当前节点集没有按升序排列，则打印“No” 。否则，检查下一次迭代。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// TreeNode class
struct TreeNode
{
int val;
TreeNode* left,*right;
TreeNode(int v)
{
val = v;
left = NULL;
right = NULL;
}
};
// Function to check if the nodes
// are in increasing order or not
bool isInc(TreeNode *root)
{
// Perform Level Order Traversal
queue que;
que.push(root);
while (true)
{
// Current length of queue
int length = que.size();
// If queue is empty
if (length == 0)
break;
auto pre = que.front();
// Level order traversal
while (length > 0)
{
// Pop element from
// front of the queue
auto temp = que.front();
que.pop();
// If previous value exceeds
// current value, return false
if (pre->val > temp->val)
return false;
pre = temp;
if (temp->left)
que.push(temp->left);
if (temp->right)
que.push(temp->right);
length -= 1;
}
}
return true;
}
// Function to print the Tree
// after modification
void levelOrder(TreeNode *root)
{
// Performs level
// order traversal
queue que;
que.push(root);
while (true)
{
// Calculate size of the queue
int length = que.size();
if (length == 0)
break;
// Iterate until queue is empty
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;
}
cout << endl;
}
// Function to arrange node values
// of each level in increasing order
void makeInc(TreeNode *root)
{
// Perform level order traversal
queue que;
que.push(root);
while (true)
{
// Calculate length of queue
int length = que.size();
// If queue is empty
if (length == 0)
break;
int prev = -1;
// Level order traversal
while (length > 0)
{
//cout<<"loop";
// Pop element from
// front of the queue
auto temp = que.front();
que.pop();
// Initialize the optimal
// element by the initial
// element
auto optEle = temp->val;
auto strEle = to_string(temp->val);
// Check for all left
// shift operations
bool flag = true;
int yy = strEle.size();
for(int idx = 0; idx < strEle.size(); idx++)
{
// Left shift
int ls = stoi(strEle.substr(idx, yy) +
strEle.substr(0, idx));
if (ls >= prev and flag)
{
optEle = ls;
flag = false;
}
// If the current shifting
// gives optimal solution
if (ls >= prev)
optEle = min(optEle, ls);
}
// Replacing initial element
// by the optimal element
temp->val = optEle;
prev = temp->val;
// Push the LST
if (temp->left)
que.push(temp->left);
// Push the RST
if (temp->right)
que.push(temp->right);
length -= 1;
}
}
// Print the result
if (isInc(root))
levelOrder(root);
else
cout << (-1);
}
// Driver Code
int main()
{
TreeNode *root = new TreeNode(341);
root->left = new TreeNode(241);
root->right = new TreeNode(123);
root->left->left = new TreeNode(324);
root->left->right = new TreeNode(235);
root->right->right = new TreeNode(161);
makeInc(root);
}
// This code is contributed by mohit kumar 29 Java
// Java program for the above approach
import java.util.*;
class GFG{
// TreeNode class
static class TreeNode
{
public int val;
public TreeNode left,right;
};
static TreeNode newNode(int v)
{
TreeNode temp = new TreeNode();
temp.val = v;
temp.left = temp.right = null;
return temp;
}
// Function to check if the nodes
// are in increasing order or not
static boolean isInc(TreeNode root)
{
// Perform Level Order Traversal
Queue que = new LinkedList<>();
que.add(root);
while (true)
{
// Current len of queue
int len = que.size();
// If queue is empty
if (len == 0)
break;
TreeNode pre = que.peek();
// Level order traversal
while (len > 0)
{
// Pop element from
// front of the queue
TreeNode temp = que.peek();
que.poll();
// If previous value exceeds
// current value, return false
if (pre.val > temp.val)
return false;
pre = temp;
if (temp.left != null)
que.add(temp.left);
if (temp.right != null)
que.add(temp.right);
len -= 1;
}
}
return true;
}
// Function to print the Tree
// after modification
static void levelOrder(TreeNode root)
{
// Performs level
// order traversal
Queue que = new LinkedList<>();
que.add(root);
while (true)
{
// Calculate size of the queue
int len = que.size();
if (len == 0)
break;
// Iterate until queue is empty
while (len > 0)
{
TreeNode temp = que.peek();
que.poll();
System.out.print(temp.val+" ");
if (temp.left != null)
que.add(temp.left);
if (temp.right != null)
que.add(temp.right);
len -= 1;
}
System.out.println();
}
System.out.println();
}
// Function to arrange node values
// of each level in increasing order
static void makeInc(TreeNode root)
{
// Perform level order traversal
Queue que = new LinkedList<>();
que.add(root);
while (true)
{
// Calculate len of queue
int len = que.size();
// If queue is empty
if (len == 0)
break;
int prev = -1;
// Level order traversal
while (len > 0)
{
//cout<<"loop";
// Pop element from
// front of the queue
TreeNode temp = que.peek();
que.poll();
// Initialize the optimal
// element by the initial
// element
int optEle = temp.val;
String strEle = String.valueOf(optEle);
// Check for all left
// shift operations
boolean flag = true;
int yy = strEle.length();
for(int idx = 0; idx < strEle.length(); idx++)
{
// Left shift
String s1 = strEle.substring(idx, yy);
String s2 = strEle.substring(0, idx);
String s = s1+ s2;
int ls = Integer.valueOf(s);
if (ls >= prev && flag)
{
optEle = ls;
flag = false;
}
// If the current shifting
// gives optimal solution
if (ls >= prev)
optEle = Math.min(optEle, ls);
}
// Replacing initial element
// by the optimal element
temp.val = optEle;
prev = temp.val;
// Push the LST
if (temp.left != null)
que.add(temp.left);
// Push the RST
if (temp.right != null)
que.add(temp.right);
len -= 1;
}
}
// Print the result
if (isInc(root) == true)
levelOrder(root);
else
System.out.print(-1);
}
// Driver code
public static void main (String[] args)
{
TreeNode root = newNode(341);
root.left = newNode(241);
root.right = newNode(123);
root.left.left = newNode(324);
root.left.right = newNode(235);
root.right.right = newNode(161);
makeInc(root);
}
}
// This code is contributed by offbeat Python3
# Python3 program for the above approach
# TreeNode class
class TreeNode:
def __init__(self, val = 0, left = None, right = None):
self.val = val
self.left = left
self.right = right
# Function to check if the nodes
# are in increasing order or not
def isInc(root):
# Perform Level Order Traversal
que = [root]
while True:
# Current length of queue
length = len(que)
# If queue is empty
if not length:
break
pre = que[0]
# Level order traversal
while length:
# Pop element from
# front of the queue
temp = que.pop(0)
# If previous value exceeds
# current value, return false
if pre.val > temp.val:
return False
pre = temp
if temp.left:
que.append(temp.left)
if temp.right:
que.append(temp.right)
length -= 1
return True
# Function to arrange node values
# of each level in increasing order
def makeInc(root):
# Perform level order traversal
que = [root]
while True:
# Calculate length of queue
length = len(que)
# If queue is empty
if not length:
break
prev = -1
# Level order traversal
while length:
# Pop element from
# front of the queue
temp = que.pop(0)
# Initialize the optimal
# element by the initial
# element
optEle = temp.val
strEle = str(temp.val)
# Check for all left
# shift operations
flag = True
for idx in range(len(strEle)):
# Left shift
ls = int(strEle[idx:] + strEle[:idx])
if ls >= prev and flag:
optEle = ls
flag = False
# If the current shifting
# gives optimal solution
if ls >= prev:
optEle = min(optEle, ls)
# Replacing initial element
# by the optimal element
temp.val = optEle
prev = temp.val
# Push the LST
if temp.left:
que.append(temp.left)
# Push the RST
if temp.right:
que.append(temp.right)
length -= 1
# Print the result
if isInc(root):
levelOrder(root)
else:
print(-1)
# Function to print the Tree
# after modification
def levelOrder(root):
# Performs level
# order traversal
que = [root]
while True:
# Calculate size of the queue
length = len(que)
if not length:
break
# Iterate until queue is empty
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
print()
# Driver Code
root = TreeNode(341)
root.left = TreeNode(241)
root.right = TreeNode(123)
root.left.left = TreeNode(324)
root.left.right = TreeNode(235)
root.right.right = TreeNode(161)
makeInc(root)C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// TreeNode class
class TreeNode
{
public int val;
public TreeNode left,right;
};
static TreeNode newNode(int v)
{
TreeNode temp = new TreeNode();
temp.val = v;
temp.left = temp.right = null;
return temp;
}
// Function to check if the nodes
// are in increasing order or not
static bool isInc(TreeNode root)
{
// Perform Level Order Traversal
Queue que = new Queue();
que.Enqueue(root);
while (true)
{
// Current len of queue
int len = que.Count;
// If queue is empty
if (len == 0)
break;
TreeNode pre = que.Peek();
// Level order traversal
while (len > 0)
{
// Pop element from
// front of the queue
TreeNode temp = que.Peek();
que.Dequeue();
// If previous value exceeds
// current value, return false
if (pre.val > temp.val)
return false;
pre = temp;
if (temp.left != null)
que.Enqueue(temp.left);
if (temp.right != null)
que.Enqueue(temp.right);
len -= 1;
}
}
return true;
}
// Function to print the Tree
// after modification
static void levelOrder(TreeNode root)
{
// Performs level
// order traversal
Queue que = new Queue();
que.Enqueue(root);
while (true)
{
// Calculate size of the queue
int len = que.Count;
if (len == 0)
break;
// Iterate until queue is empty
while (len > 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);
len -= 1;
}
Console.Write("\n");
}
Console.Write("\n");
}
// Function to arrange node values
// of each level in increasing order
static void makeInc(TreeNode root)
{
// Perform level order traversal
Queue que = new Queue();
que.Enqueue(root);
while (true)
{
// Calculate len of queue
int len = que.Count;
// If queue is empty
if (len == 0)
break;
int prev = -1;
// Level order traversal
while (len > 0)
{
//cout<<"loop";
// Pop element from
// front of the queue
TreeNode temp = que.Peek();
que.Dequeue();
// Initialize the optimal
// element by the initial
// element
int optEle = temp.val;
string strEle = optEle.ToString();
// Check for all left
// shift operations
bool flag = true;
int yy = strEle.Length;
for(int idx = 0; idx < strEle.Length; idx++)
{
// Left shift
string s1 = strEle.Substring(idx, yy - idx);
string s2 = strEle.Substring(0, idx);
string s = String.Concat(s1, s2);
int ls = Int32.Parse(s);
if (ls >= prev && flag)
{
optEle = ls;
flag = false;
}
// If the current shifting
// gives optimal solution
if (ls >= prev)
optEle = Math.Min(optEle, ls);
}
// Replacing initial element
// by the optimal element
temp.val = optEle;
prev = temp.val;
// Push the LST
if (temp.left != null)
que.Enqueue(temp.left);
// Push the RST
if (temp.right != null)
que.Enqueue(temp.right);
len -= 1;
}
}
// Print the result
if (isInc(root) == true)
levelOrder(root);
else
Console.Write(-1);
}
// Driver Code
public static void Main()
{
TreeNode root = newNode(341);
root.left = newNode(241);
root.right = newNode(123);
root.left.left = newNode(324);
root.left.right = newNode(235);
root.right.right = newNode(161);
makeInc(root);
}
}
// This code is contributed by ipg2016107 输出：
134
124 231
243 352 611时间复杂度： O(N)
辅助空间： O(N)
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