从给定的中序遍历构造特殊二叉树
给定一个特殊二叉树的中序遍历，其中每个节点的键大于左右子节点的键，构造二叉树并返回根。
例子：
Input: inorder[] = {5, 10, 40, 30, 28}
Output: root of following tree
40
/ \
10 30
/ \
5 28
Input: inorder[] = {1, 5, 10, 40, 30,
15, 28, 20}
Output: root of following tree
40
/ \
10 30
/ \
5 28
/ / \
1 15 20从给定的 Inorder 和 Preorder 遍历构造树中使用的想法可以在这里使用。让给定的数组是 {1, 5, 10, 40, 30, 15, 28, 20}。给定数组中的最大元素必须是根。最大元素左侧的元素在左子树中，右侧的元素在右子树中。
40
/ \
{1,5,10} {30,15,28,20}我们对左右子树递归地按照上面的步骤，最终得到下面的树。
40
/ \
10 30
/ \
5 28
/ / \
1 15 20算法： buildTree()
1) 查找数组中最大元素的索引。最大元素必须是二叉树的根。
2）创建一个新的树节点“根”，其数据为步骤 1 中找到的最大值。
3) 为最大元素之前的元素调用buildTree，并将构建的树作为'root'的左子树。
5) 为最大元素之后的元素调用buildTree，并将构建的树作为'root'的右子树。
6）返回“根”。
实现：以下是上述算法的实现。
C++
/* C++ program to construct tree
from inorder traversal */
#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;
node* right;
};
/* Prototypes of a utility function to get the maximum
value in inorder[start..end] */
int max(int inorder[], int strt, int end);
/* A utility function to allocate memory for a node */
node* newNode(int data);
/* Recursive function to construct binary of size len from
Inorder traversal inorder[]. Initial values of start and end
should be 0 and len -1. */
node* buildTree (int inorder[], int start, int end)
{
if (start > end)
return NULL;
/* Find index of the maximum element from Binary Tree */
int i = max (inorder, start, end);
/* Pick the maximum value and make it root */
node *root = newNode(inorder[i]);
/* If this is the only element in inorder[start..end],
then return it */
if (start == end)
return root;
/* Using index in Inorder traversal, construct left and
right subtress */
root->left = buildTree (inorder, start, i - 1);
root->right = buildTree (inorder, i + 1, end);
return root;
}
/* UTILITY FUNCTIONS */
/* Function to find index of the maximum value in arr[start...end] */
int max (int arr[], int strt, int end)
{
int i, max = arr[strt], maxind = strt;
for(i = strt + 1; i <= end; i++)
{
if(arr[i] > max)
{
max = arr[i];
maxind = i;
}
}
return maxind;
}
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
node* newNode (int data)
{
node* Node = new node();
Node->data = data;
Node->left = NULL;
Node->right = NULL;
return Node;
}
/* This function is here just to test buildTree() */
void printInorder (node* node)
{
if (node == NULL)
return;
/* first recur on left child */
printInorder (node->left);
/* then print the data of node */
cout<data<<" ";
/* now recur on right child */
printInorder (node->right);
}
/* Driver code*/
int main()
{
/* Assume that inorder traversal of following tree is given
40
/ \
10 30
/ \
5 28 */
int inorder[] = {5, 10, 40, 30, 28};
int len = sizeof(inorder)/sizeof(inorder[0]);
node *root = buildTree(inorder, 0, len - 1);
/* Let us test the built tree by printing Inorder traversal */
cout << "Inorder traversal of the constructed tree is \n";
printInorder(root);
return 0;
}
// This is code is contributed by rathbhupendra C
/* program to construct tree from inorder traversal */
#include
#include
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct node
{
int data;
struct node* left;
struct node* right;
};
/* Prototypes of a utility function to get the maximum
value in inorder[start..end] */
int max(int inorder[], int strt, int end);
/* A utility function to allocate memory for a node */
struct node* newNode(int data);
/* Recursive function to construct binary of size len from
Inorder traversal inorder[]. Initial values of start and end
should be 0 and len -1. */
struct node* buildTree (int inorder[], int start, int end)
{
if (start > end)
return NULL;
/* Find index of the maximum element from Binary Tree */
int i = max (inorder, start, end);
/* Pick the maximum value and make it root */
struct node *root = newNode(inorder[i]);
/* If this is the only element in inorder[start..end],
then return it */
if (start == end)
return root;
/* Using index in Inorder traversal, construct left and
right subtress */
root->left = buildTree (inorder, start, i-1);
root->right = buildTree (inorder, i+1, end);
return root;
}
/* UTILITY FUNCTIONS */
/* Function to find index of the maximum value in arr[start...end] */
int max (int arr[], int strt, int end)
{
int i, max = arr[strt], maxind = strt;
for(i = strt+1; i <= end; i++)
{
if(arr[i] > max)
{
max = arr[i];
maxind = i;
}
}
return maxind;
}
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
struct node* newNode (int data)
{
struct node* node = (struct node*)malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return node;
}
/* This function is here just to test buildTree() */
void printInorder (struct node* node)
{
if (node == NULL)
return;
/* first recur on left child */
printInorder (node->left);
/* then print the data of node */
printf("%d ", node->data);
/* now recur on right child */
printInorder (node->right);
}
/* Driver program to test above functions */
int main()
{
/* Assume that inorder traversal of following tree is given
40
/ \
10 30
/ \
5 28 */
int inorder[] = {5, 10, 40, 30, 28};
int len = sizeof(inorder)/sizeof(inorder[0]);
struct node *root = buildTree(inorder, 0, len - 1);
/* Let us test the built tree by printing Inorder traversal */
printf("\n Inorder traversal of the constructed tree is \n");
printInorder(root);
return 0;
} Java
// Java program to construct tree from inorder traversal
/* A binary tree node has data, pointer to left child
and a pointer to right child */
class Node
{
int data;
Node left, right;
Node(int item)
{
data = item;
left = right = null;
}
}
class BinaryTree
{
Node root;
/* Recursive function to construct binary of size len from
Inorder traversal inorder[]. Initial values of start and end
should be 0 and len -1. */
Node buildTree(int inorder[], int start, int end, Node node)
{
if (start > end)
return null;
/* Find index of the maximum element from Binary Tree */
int i = max(inorder, start, end);
/* Pick the maximum value and make it root */
node = new Node(inorder[i]);
/* If this is the only element in inorder[start..end],
then return it */
if (start == end)
return node;
/* Using index in Inorder traversal, construct left and
right subtress */
node.left = buildTree(inorder, start, i - 1, node.left);
node.right = buildTree(inorder, i + 1, end, node.right);
return node;
}
/* UTILITY FUNCTIONS */
/* Function to find index of the maximum value in arr[start...end] */
int max(int arr[], int strt, int end)
{
int i, max = arr[strt], maxind = strt;
for (i = strt + 1; i <= end; i++)
{
if (arr[i] > max)
{
max = arr[i];
maxind = i;
}
}
return maxind;
}
/* This function is here just to test buildTree() */
void printInorder(Node node)
{
if (node == null)
return;
/* first recur on left child */
printInorder(node.left);
/* then print the data of node */
System.out.print(node.data + " ");
/* now recur on right child */
printInorder(node.right);
}
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
/* Assume that inorder traversal of following tree is given
40
/ \
10 30
/ \
5 28 */
int inorder[] = new int[]{5, 10, 40, 30, 28};
int len = inorder.length;
Node mynode = tree.buildTree(inorder, 0, len - 1, tree.root);
/* Let us test the built tree by printing Inorder traversal */
System.out.println("Inorder traversal of the constructed tree is ");
tree.printInorder(mynode);
}
}
// This code has been contributed by Mayank JaiswalPython3
# Python3 program to construct tree from
# inorder traversal
# Helper class that allocates a new node
# with the given data and None left and
# right pointers.
class newNode:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Prototypes of a utility function to get
# the maximum value in inorder[start..end]
# Recursive function to construct binary of
# size len from Inorder traversal inorder[].
# Initial values of start and end should be
# 0 and len -1.
def buildTree (inorder, start, end):
if start > end:
return None
# Find index of the maximum element
# from Binary Tree
i = Max (inorder, start, end)
# Pick the maximum value and make it root
root = newNode(inorder[i])
# If this is the only element in
# inorder[start..end], then return it
if start == end:
return root
# Using index in Inorder traversal,
# construct left and right subtress
root.left = buildTree (inorder, start, i - 1)
root.right = buildTree (inorder, i + 1, end)
return root
# UTILITY FUNCTIONS
# Function to find index of the maximum
# value in arr[start...end]
def Max(arr, strt, end):
i, Max = 0, arr[strt]
maxind = strt
for i in range(strt + 1, end + 1):
if arr[i] > Max:
Max = arr[i]
maxind = i
return maxind
# This function is here just to test buildTree()
def printInorder (node):
if node == None:
return
# first recur on left child
printInorder (node.left)
# then print the data of node
print(node.data, end = " ")
# now recur on right child
printInorder (node.right)
# Driver Code
if __name__ == '__main__':
# Assume that inorder traversal of
# following tree is given
# 40
# / \
# 10 30
# / \
#5 28
inorder = [5, 10, 40, 30, 28]
Len = len(inorder)
root = buildTree(inorder, 0, Len - 1)
# Let us test the built tree by
# printing Inorder traversal
print("Inorder traversal of the",
"constructed tree is ")
printInorder(root)
# This code is contributed by PranchalKC#
// C# program to construct tree
// from inorder traversal
using System;
/* A binary tree node has data,
pointer to left child and a pointer
to right child */
public class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
class GFG
{
public Node root;
/* Recursive function to construct binary
of size len from Inorder traversal inorder[].
Initial values of start and end should be 0
and len -1. */
public virtual Node buildTree(int[] inorder, int start,
int end, Node node)
{
if (start > end)
{
return null;
}
/* Find index of the maximum
element from Binary Tree */
int i = max(inorder, start, end);
/* Pick the maximum value and
make it root */
node = new Node(inorder[i]);
/* If this is the only element in
inorder[start..end], then return it */
if (start == end)
{
return node;
}
/* Using index in Inorder traversal,
construct left and right subtress */
node.left = buildTree(inorder, start,
i - 1, node.left);
node.right = buildTree(inorder, i + 1,
end, node.right);
return node;
}
/* UTILITY FUNCTIONS */
/* Function to find index of the
maximum value in arr[start...end] */
public virtual int max(int[] arr,
int strt, int end)
{
int i, max = arr[strt], maxind = strt;
for (i = strt + 1; i <= end; i++)
{
if (arr[i] > max)
{
max = arr[i];
maxind = i;
}
}
return maxind;
}
/* This function is here just
to test buildTree() */
public virtual void printInorder(Node node)
{
if (node == null)
{
return;
}
/* first recur on left child */
printInorder(node.left);
/* then print the data of node */
Console.Write(node.data + " ");
/* now recur on right child */
printInorder(node.right);
}
// Driver Code
public static void Main(string[] args)
{
GFG tree = new GFG();
/* Assume that inorder traversal
of following tree is given
40
/ \
10 30
/ \
5 28 */
int[] inorder = new int[]{5, 10, 40, 30, 28};
int len = inorder.Length;
Node mynode = tree.buildTree(inorder, 0,
len - 1, tree.root);
/* Let us test the built tree by
printing Inorder traversal */
Console.WriteLine("Inorder traversal of " +
"the constructed tree is ");
tree.printInorder(mynode);
}
}
// This code is contributed by Shrikant13Javascript
输出：
Inorder traversal of the constructed tree is
5 10 40 30 28时间复杂度： O(n^2)