给定一棵二叉树，任务是将给定的二叉树转换为最小-最大乘积树，并打印修改后的树的级序序列。
Min-Max Product Tree： Min-Max Product Tree 包含其每个节点的左右子树的最小值和最大值的乘积。
注意：对于任何只有一个子节点的节点，该子节点的值将被视为最小值和最大值。
例子：
Input:
1
/ \
12 11
/ / \
3 4 13
\ /
15 5
Output:
45 9 60 3 225 25 15 5
Explanation:
Min-Max Product Tree:
45 (3 * 15)
/ \
( 3 * 3)9 60 (4 * 15)
/ / \
3 225 25 (5* 5)
\ /
15 5
Input:
5
/ \
21 77
/ \ \
61 16 16
\ /
10 3
/
23
Output:
231 610 48 61 230 9 529 3 23方法：
思路是利用后序遍历递归遍历左右子树，提取最小值和最大值。将每个节点的最小值和最大值的乘积存储在 Min-Max Product 树中。计算一次，将当前节点值与当前最小值和最大值进行比较并进行相应修改。在更高级别返回节点的新最小值和最大值。对所有节点重复此过程，最后打印 Min-Max Product 树的层序遍历。
下面是上述方法的实现：
C++
// C++ implementation of
// the above approach
#include
using namespace std;
// A Tree node
struct Node {
int data;
struct Node *left, *right;
};
// Utility function to create
// a new node
Node* newNode(int key)
{
Node* temp = new Node;
temp->data = key;
temp->left = temp->right = NULL;
return (temp);
}
// A minMax Structure for storing
// minimum and maximum value
struct minMax {
int min;
int max;
};
// Function to return min value
int min(minMax* a, minMax* b)
{
if (a != NULL
&& b != NULL) {
return a->min < b->min
? a->min
: b->min;
}
return a == NULL
? b->min
: a->min;
}
// Function to return max value
int max(minMax* a, minMax* b)
{
if (a != NULL
&& b != NULL) {
return a->max > b->max
? a->max
: b->max;
}
return a == NULL
? b->max
: a->max;
}
// Function to return min value
int min(int x, int y)
{
return x < y ? x : y;
}
// Function to return max value
int max(int x, int y)
{
return x > y ? x : y;
}
// Utility function to create
// min-max product tree
minMax* minMaxTreeUtil(
Node* root)
{
// Base condition
if (root == NULL) {
return NULL;
}
minMax* var = new minMax;
// Condition to check if
// current node is leaf node
if (root->left == NULL
&& root->right == NULL) {
var->min = root->data;
var->max = root->data;
return var;
}
// Left recursive call
minMax* left
= minMaxTreeUtil(root->left);
// Right recursive call
minMax* right
= minMaxTreeUtil(root->right);
// Store Min
var->min = min(left, right);
// Store Max
var->max = max(left, right);
// Store current node data
int currData = root->data;
// Assign product of minimum
// and maximum value
root->data = var->min * var->max;
// Again store min by considering
// current node value
var->min = min(var->min, currData);
// Again store max by considering
// current node value
var->max = max(var->max, currData);
return var;
}
void print(Node* root)
{
// Base Case
if (root == NULL)
return;
// Create an empty queue for
// level order traversal
queue q;
// Enqueue Root and initialize
// height
q.push(root);
while (q.empty() == false) {
// nodeCount (queue size)
// indicates number
// of nodes at current level.
int nodeCount = q.size();
// Dequeue all nodes of current
// level and Enqueue all nodes
// of next level
while (nodeCount > 0) {
Node* node = q.front();
cout << node->data << " ";
q.pop();
if (node->left != NULL)
q.push(node->left);
if (node->right != NULL)
q.push(node->right);
nodeCount--;
}
}
}
void minMaxProductTree(Node* root)
{
// Utility Function call
minMaxTreeUtil(root);
// Print tree
print(root);
}
// Driver Code
int main()
{
/* 10
/ \
48 3
/ \
11 37
/ \ / \
7 29 42 19
/
7
*/
// Create Binary Tree as shown
Node* root = newNode(10);
root->left = newNode(48);
root->right = newNode(3);
root->right->left = newNode(11);
root->right->right = newNode(37);
root->right->left->left = newNode(7);
root->right->left->right = newNode(29);
root->right->right->left = newNode(42);
root->right->right->right = newNode(19);
root->right->right->right->left = newNode(7);
// Create Min Max Product Tree
minMaxProductTree(root);
return 0;
} Python3
# Python3 program for the
# above approach
from collections import deque
# A Tree node
class Node:
def __init__(self, x):
self.data = x
self.left = None
self.right = None
# A minMax Structure for storing
# minimum and maximum value
class minMax:
def __init__(self, x, y):
self.min = x
self.max = y
# Function to return min value
def minm(a, b):
if (a != None and
b != None):
if a.min < b.min:
return a.min
return b.min
if a == None:
return b.min
return a.min
# Function to return max value
def maxm(a, b):
if a != None and b != None:
if a.max > b.max:
return a.max
return b.max
if a == None:
return b.max
return a.max
# Utility function to create
# min-max product tree
def minMaxTreeUtil(root):
# Base condition
if (root == None):
return None
var = minMax(10 ** 9,
-10 ** 9)
# Condition to check if
# current node is leaf node
if (root.left == None and
root.right == None):
var.min = root.data
var.max = root.data
return var
# Left recursive call
left= minMaxTreeUtil(root.left)
# Right recursive call
right= minMaxTreeUtil(root.right)
# Store Min
var.min = minm(left, right)
# Store Max
var.max = maxm(left, right)
# Store current node data
currData = root.data
# Assign product of minimum
# and maximum value
root.data = var.min * var.max
# Again store min by considering
# current node value
var.min = min(var.min,
currData)
# Again store max by considering
# current node value
var.max = max(var.max,
currData)
return var
def printt(root):
# Base Case
if (root == None):
return
# Create an empty queue for
# level order traversal
q = deque()
# Enqueue Root and initialize
# height
q.append(root)
while (len(q) > 0):
# nodeCount (queue size)
# indicates number
# of nodes at current level.
nodeCount = len(q)
# Dequeue all nodes of current
# level and Enqueue all nodes
# of next level
while (nodeCount > 0):
node = q.popleft()
print(node.data,
end = " ")
if (node.left != None):
q.append(node.left)
if (node.right != None):
q.append(node.right)
nodeCount -= 1
def minMaxProductTree(root):
# Utility Function call
minMaxTreeUtil(root)
# Prtree
printt(root)
# Driver Code
if __name__ == '__main__':
# /* 10
# / \
# 48 3
# / \
# 11 37
# / \ / \
# 7 29 42 19
# /
# 7
# */
# Create Binary Tree as shown
root = Node(10)
root.left = Node(48)
root.right = Node(3)
root.right.left = Node(11)
root.right.right = Node(37)
root.right.left.left = Node(7)
root.right.left.right = Node(29)
root.right.right.left = Node(42)
root.right.right.right = Node(19)
root.right.right.right.left = Node(7)
# Create Min Max Product Tree
minMaxProductTree(root)
# This code is contributed by Mohit Kumar 29输出：
144 48 294 203 294 7 29 42 49 7时间复杂度： O(N)，其中 N 表示二叉树中的节点数
辅助空间： O(1)
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