给定两个二叉树，任务是从两个给定的二叉树中创建一个最大二叉树，并打印该树的中序遍历。
什么是最大的二叉树?
The maximum binary is constructed in the following manner:
In the case of both the Binary Trees having two corresponding nodes, the maximum of the two values is considered as the node value of the Maximum Binary Tree.
If any of the two nodes is NULL and if the other node is not null, insert that value on that node of the Maximum Binary Tree.
例子：
Input:
Tree 1 Tree 2
3 5
/ \ / \
2 6 1 8
/ \ \
20 2 8
Output: 20 2 2 5 8 8
Explanation:
5
/ \
2 8
/ \ \
20 2 8
To construct the required Binary Tree,
Root Node value: Max(3, 5) = 5
Root->left value: Max(2, 1) = 2
Root->right value: Max(6, 8) = 8
Root->left->left value: 20
Root->left->right value: 2
Root->right->right value: 8
Input:
Tree 1 Tree 2
9 5
/ \ / \
2 6 1 8
/ \ \ \
20 3 2 8
Output: 20 2 3 9 8 8
Explanation:
9
/ \
2 8
/ \ \
20 3 8方法：
请按照以下步骤解决问题：
- 使用preorder traversal遍历两棵树。
- 如果两个节点都是NULL ，则返回。否则，请检查以下条件：
- 如果两个节点都不为NULL，则将它们之间的最大值存储为最大二叉树的节点值。
- 如果只有一个节点为NULL ，则将非 NULL节点的值存储为最大二叉树的节点值。
- 递归遍历左子树。
- 递归遍历右子树
- 最后，返回最大二叉树的根。
下面是上述方法的实现：
C++
// C++ program to find the Maximum
// Binary Tree from two Binary Trees
#include
using namespace std;
// A binary tree node has data,
// pointer to left child
// and a pointer to right child
class Node{
public:
int data;
Node *left, *right;
Node(int data, Node *left,
Node *right)
{
this->data = data;
this->left = left;
this->right = right;
}
};
// Helper method that allocates
// a new node with the given data
// and NULL left and right pointers.
Node* newNode(int data)
{
Node *tmp = new Node(data, NULL, NULL);
return tmp;
}
// Given a binary tree, print
// its nodes in inorder
void inorder(Node *node)
{
if (node == NULL)
return;
// First recur on left child
inorder(node->left);
// Then print the data of node
cout << node->data << " ";
// Now recur on right child
inorder(node->right);
}
// Method to find the maximum
// binary tree from two binary trees
Node* MaximumBinaryTree(Node *t1, Node *t2)
{
if (t1 == NULL)
return t2;
if (t2 == NULL)
return t1;
t1->data = max(t1->data, t2->data);
t1->left = MaximumBinaryTree(t1->left,
t2->left);
t1->right = MaximumBinaryTree(t1->right,
t2->right);
return t1;
}
// Driver Code
int main()
{
/* First Binary Tree
3
/ \
2 6
/
20
*/
Node *root1 = newNode(3);
root1->left = newNode(2);
root1->right = newNode(6);
root1->left->left = newNode(20);
/* Second Binary Tree
5
/ \
1 8
\ \
2 8
*/
Node *root2 = newNode(5);
root2->left = newNode(1);
root2->right = newNode(8);
root2->left->right = newNode(2);
root2->right->right = newNode(8);
Node *root3 = MaximumBinaryTree(root1, root2);
inorder(root3);
}
// This code is contributed by pratham76 Java
// Java program to find the Maximum
// Binary Tree from two Binary Trees
/* A binary tree node has data,
pointer to left child
and a pointer to right child */
class Node {
int data;
Node left, right;
public Node(int data, Node left,
Node right)
{
this.data = data;
this.left = left;
this.right = right;
}
/* Helper method that allocates
a new node with the given data
and NULL left and right pointers. */
static Node newNode(int data)
{
return new Node(data, null, null);
}
/* Given a binary tree, print
its nodes in inorder*/
static void inorder(Node node)
{
if (node == null)
return;
/* first recur on left child */
inorder(node.left);
/* then print the data of node */
System.out.printf("%d ", node.data);
/* now recur on right child */
inorder(node.right);
}
/* Method to find the maximum
binary tree from
two binary trees*/
static Node MaximumBinaryTree(Node t1, Node t2)
{
if (t1 == null)
return t2;
if (t2 == null)
return t1;
t1.data = Math.max(t1.data, t2.data);
t1.left = MaximumBinaryTree(t1.left,
t2.left);
t1.right = MaximumBinaryTree(t1.right,
t2.right);
return t1;
}
// Driver Code
public static void main(String[] args)
{
/* First Binary Tree
3
/ \
2 6
/
20
*/
Node root1 = newNode(3);
root1.left = newNode(2);
root1.right = newNode(6);
root1.left.left = newNode(20);
/* Second Binary Tree
5
/ \
1 8
\ \
2 8
*/
Node root2 = newNode(5);
root2.left = newNode(1);
root2.right = newNode(8);
root2.left.right = newNode(2);
root2.right.right = newNode(8);
Node root3
= MaximumBinaryTree(root1, root2);
inorder(root3);
}
}Python3
# Python3 program to find the Maximum
# Binary Tree from two Binary Trees
''' A binary tree node has data,
pointer to left child
and a pointer to right child '''
class Node:
def __init__(self, data, left, right):
self.data = data
self.left = left
self.right = right
''' Helper method that allocates
a new node with the given data
and None left and right pointers. '''
def newNode(data):
return Node(data, None, None);
''' Given a binary tree, print
its nodes in inorder'''
def inorder(node):
if (node == None):
return;
''' first recur on left child '''
inorder(node.left);
''' then print the data of node '''
print(node.data, end=' ');
''' now recur on right child '''
inorder(node.right);
''' Method to find the maximum
binary tree from
two binary trees'''
def MaximumBinaryTree(t1, t2):
if (t1 == None):
return t2;
if (t2 == None):
return t1;
t1.data = max(t1.data, t2.data);
t1.left = MaximumBinaryTree(t1.left,
t2.left);
t1.right = MaximumBinaryTree(t1.right,
t2.right);
return t1;
# Driver Code
if __name__=='__main__':
''' First Binary Tree
3
/ \
2 6
/
20
'''
root1 = newNode(3);
root1.left = newNode(2);
root1.right = newNode(6);
root1.left.left = newNode(20);
''' Second Binary Tree
5
/ \
1 8
\ \
2 8
'''
root2 = newNode(5);
root2.left = newNode(1);
root2.right = newNode(8);
root2.left.right = newNode(2);
root2.right.right = newNode(8);
root3 = MaximumBinaryTree(root1, root2);
inorder(root3);
# This code is contributed by rutvik_56C#
// C# program to find the Maximum
// Binary Tree from two Binary Trees
using System;
// A binary tree node has data,
// pointer to left child
// and a pointer to right child
class Node{
public int data;
public Node left, right;
public Node(int data, Node left,
Node right)
{
this.data = data;
this.left = left;
this.right = right;
}
// Helper method that allocates
// a new node with the given data
// and NULL left and right pointers.
static Node newNode(int data)
{
return new Node(data, null, null);
}
// Given a binary tree, print
// its nodes in inorder
static void inorder(Node node)
{
if (node == null)
return;
// first recur on left child
inorder(node.left);
// then print the data of node
Console.Write("{0} ", node.data);
// now recur on right child
inorder(node.right);
}
// Method to find the maximum
// binary tree from
// two binary trees
static Node MaximumBinaryTree(Node t1, Node t2)
{
if (t1 == null)
return t2;
if (t2 == null)
return t1;
t1.data = Math.Max(t1.data, t2.data);
t1.left = MaximumBinaryTree(t1.left,
t2.left);
t1.right = MaximumBinaryTree(t1.right,
t2.right);
return t1;
}
// Driver Code
public static void Main(String[] args)
{
/* First Binary Tree
3
/ \
2 6
/
20
*/
Node root1 = newNode(3);
root1.left = newNode(2);
root1.right = newNode(6);
root1.left.left = newNode(20);
/* Second Binary Tree
5
/ \
1 8
\ \
2 8
*/
Node root2 = newNode(5);
root2.left = newNode(1);
root2.right = newNode(8);
root2.left.right = newNode(2);
root2.right.right = newNode(8);
Node root3 = MaximumBinaryTree(root1, root2);
inorder(root3);
}
}
// This code is contributed by Amal Kumar Choubey输出：
20 2 2 5 8 8时间复杂度： O(N)
辅助空间： O(N)
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