Construct a Binary Search Tree from given postorder
Last Updated :
20 Feb, 2023
Given postorder traversal of a binary search tree, construct the BST.
For example, if the given traversal is {1, 7, 5, 50, 40, 10}, then following tree should be constructed and root of the tree should be returned.
10
/ \
5 40
/ \ \
1 7 50
Method 1 ( O(n^2) time complexity ):
The last element of postorder traversal is always root. We first construct the root. Then we find the index of last element which is smaller than root. Let the index be 'i'. The values between 0 and 'i' are part of left subtree, and the values between 'i+1' and 'n-2' are part of right subtree. Divide given post[] at index "i" and recur for left and right sub-trees.
For example in {1, 7, 5, 50, 40, 10}, 10 is the last element, so we make it root. Now we look for the last element smaller than 10, we find 5. So we know the structure of BST is as following.
10
/ \
/ \
{1, 7, 5} {50, 40}
We recursively follow above steps for subarrays {1, 7, 5} and {40, 50}, and get the complete tree.
Method 2 ( O(n) time complexity ):
The trick is to set a range {min .. max} for every node. Initialize the range as {INT_MIN .. INT_MAX}. The last node will definitely be in range, so create root node. To construct the left subtree, set the range as {INT_MIN …root->data}. If a values is in the range {INT_MIN .. root->data}, the values is part of left subtree. To construct the right subtree, set the range as {root->data .. INT_MAX}.
Following code is used to generate the exact Binary Search Tree of a given post order traversal.
C++
/* A O(n) program for construction of
BST from postorder traversal */
#include <bits/stdc++.h>
using namespace std;
/* A binary tree node has data,
pointer to left child and a
pointer to right child */
struct node
{
int data;
struct node *left, *right;
};
// A utility function to create a node
struct node* newNode (int data)
{
struct node* temp =
(struct node *) malloc(sizeof(struct node));
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}
// A recursive function to construct
// BST from post[]. postIndex is used
// to keep track of index in post[].
struct node* constructTreeUtil(int post[], int* postIndex,
int key, int min, int max,
int size)
{
// Base case
if (*postIndex < 0)
return NULL;
struct node* root = NULL;
// If current element of post[] is
// in range, then only it is part
// of current subtree
if (key > min && key < max)
{
// Allocate memory for root of this
// subtree and decrement *postIndex
root = newNode(key);
*postIndex = *postIndex - 1;
if (*postIndex >= 0)
{
// All nodes which are in range {key..max}
// will go in right subtree, and first such
// node will be root of right subtree.
root->right = constructTreeUtil(post, postIndex,
post[*postIndex],
key, max, size );
// Construct the subtree under root
// All nodes which are in range {min .. key}
// will go in left subtree, and first such
// node will be root of left subtree.
root->left = constructTreeUtil(post, postIndex,
post[*postIndex],
min, key, size );
}
}
return root;
}
// The main function to construct BST
// from given postorder traversal.
// This function mainly uses constructTreeUtil()
struct node *constructTree (int post[],
int size)
{
int postIndex = size-1;
return constructTreeUtil(post, &postIndex,
post[postIndex],
INT_MIN, INT_MAX, size);
}
// A utility function to print
// inorder traversal of a Binary Tree
void printInorder (struct node* node)
{
if (node == NULL)
return;
printInorder(node->left);
cout << node->data << " ";
printInorder(node->right);
}
// Driver Code
int main ()
{
int post[] = {1, 7, 5, 50, 40, 10};
int size = sizeof(post) / sizeof(post[0]);
struct node *root = constructTree(post, size);
cout << "Inorder traversal of "
<< "the constructed tree: \n";
printInorder(root);
return 0;
}
// This code is contributed
// by Akanksha Rai
/* A O(n) program for construction of BST from
postorder traversal */
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct node
{
int data;
struct node *left, *right;
};
// A utility function to create a node
struct node* newNode (int data)
{
struct node* temp =
(struct node *) malloc( sizeof(struct node));
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}
// A recursive function to construct BST from post[].
// postIndex is used to keep track of index in post[].
struct node* constructTreeUtil(int post[], int* postIndex,
int key, int min, int max, int size)
{
// Base case
if (*postIndex < 0)
return NULL;
struct node* root = NULL;
// If current element of post[] is in range, then
// only it is part of current subtree
if (key > min && key < max)
{
// Allocate memory for root of this subtree and decrement
// *postIndex
root = newNode(key);
*postIndex = *postIndex - 1;
if (*postIndex >= 0)
{
// All nodes which are in range {key..max} will go in right
// subtree, and first such node will be root of right subtree.
root->right = constructTreeUtil(post, postIndex, post[*postIndex],
key, max, size );
// Construct the subtree under root
// All nodes which are in range {min .. key} will go in left
// subtree, and first such node will be root of left subtree.
root->left = constructTreeUtil(post, postIndex, post[*postIndex],
min, key, size );
}
}
return root;
}
// The main function to construct BST from given postorder
// traversal. This function mainly uses constructTreeUtil()
struct node *constructTree (int post[], int size)
{
int postIndex = size-1;
return constructTreeUtil(post, &postIndex, post[postIndex],
INT_MIN, INT_MAX, size);
}
// A utility function to print inorder traversal of a Binary Tree
void printInorder (struct node* node)
{
if (node == NULL)
return;
printInorder(node->left);
printf("%d ", node->data);
printInorder(node->right);
}
// Driver program to test above functions
int main ()
{
int post[] = {1, 7, 5, 50, 40, 10};
int size = sizeof(post) / sizeof(post[0]);
struct node *root = constructTree(post, size);
printf("Inorder traversal of the constructed tree: \n");
printInorder(root);
return 0;
}
/* A O(n) program for construction of BST from
postorder 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 data)
{
this.data = data;
left = right = null;
}
}
// Class containing variable that keeps a track of overall
// calculated postindex
class Index
{
int postindex = 0;
}
class BinaryTree
{
// A recursive function to construct BST from post[].
// postIndex is used to keep track of index in post[].
Node constructTreeUtil(int post[], Index postIndex,
int key, int min, int max, int size)
{
// Base case
if (postIndex.postindex < 0)
return null;
Node root = null;
// If current element of post[] is in range, then
// only it is part of current subtree
if (key > min && key < max)
{
// Allocate memory for root of this subtree and decrement
// *postIndex
root = new Node(key);
postIndex.postindex = postIndex.postindex - 1;
if (postIndex.postindex >= 0)
{
// All nodes which are in range {key..max} will go in
// right subtree, and first such node will be root of right
// subtree
root.right = constructTreeUtil(post, postIndex,
post[postIndex.postindex],key, max, size);
// Construct the subtree under root
// All nodes which are in range {min .. key} will go in left
// subtree, and first such node will be root of left subtree.
root.left = constructTreeUtil(post, postIndex,
post[postIndex.postindex],min, key, size);
}
}
return root;
}
// The main function to construct BST from given postorder
// traversal. This function mainly uses constructTreeUtil()
Node constructTree(int post[], int size)
{
Index index = new Index();
index.postindex = size - 1;
return constructTreeUtil(post, index, post[index.postindex],
Integer.MIN_VALUE, Integer.MAX_VALUE, size);
}
// A utility function to print inorder traversal of a Binary Tree
void printInorder(Node node)
{
if (node == null)
return;
printInorder(node.left);
System.out.print(node.data + " ");
printInorder(node.right);
}
// Driver program to test above functions
public static void main(String[] args)
{
BinaryTree tree = new BinaryTree();
int post[] = new int[]{1, 7, 5, 50, 40, 10};
int size = post.length;
Node root = tree.constructTree(post, size);
System.out.println("Inorder traversal of the constructed tree:");
tree.printInorder(root);
}
}
// This code has been contributed by Mayank Jaiswal
# A O(n) program for construction of BST
# from postorder traversal
INT_MIN = -2**31
INT_MAX = 2**31
# A binary tree node has data, pointer to
# left child and a pointer to right child
# A utility function to create a node
class newNode:
def __init__(self, data):
self.data = data
self.left = self.right = None
# A recursive function to construct
# BST from post[]. postIndex is used
# to keep track of index in post[].
def constructTreeUtil(post, postIndex,
key, min, max, size):
# Base case
if (postIndex[0] < 0):
return None
root = None
# If current element of post[] is
# in range, then only it is part
# of current subtree
if (key > min and key < max) :
# Allocate memory for root of this
# subtree and decrement *postIndex
root = newNode(key)
postIndex[0] = postIndex[0] - 1
if (postIndex[0] >= 0) :
# All nodes which are in range key..
# max will go in right subtree, and
# first such node will be root of
# right subtree.
root.right = constructTreeUtil(post, postIndex,
post[postIndex[0]],
key, max, size )
# Construct the subtree under root
# All nodes which are in range min ..
# key will go in left subtree, and
# first such node will be root of
# left subtree.
root.left = constructTreeUtil(post, postIndex,
post[postIndex[0]],
min, key, size )
return root
# The main function to construct BST
# from given postorder traversal. This
# function mainly uses constructTreeUtil()
def constructTree (post, size) :
postIndex = [size-1]
return constructTreeUtil(post, postIndex,
post[postIndex[0]],
INT_MIN, INT_MAX, size)
# A utility function to printInorder
# traversal of a Binary Tree
def printInorder (node) :
if (node == None) :
return
printInorder(node.left)
print(node.data, end = " ")
printInorder(node.right)
# Driver Code
if __name__ == '__main__':
post = [1, 7, 5, 50, 40, 10]
size = len(post)
root = constructTree(post, size)
print("Inorder traversal of the",
"constructed tree: ")
printInorder(root)
# This code is contributed
# by SHUBHAMSINGH10
using System;
/* A O(n) program for
construction of BST from
postorder traversal */
/* 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)
{
this.data = data;
left = right = null;
}
}
// Class containing variable
// that keeps a track of overall
// calculated postindex
class Index
{
public int postindex = 0;
}
public class BinaryTree
{
// A recursive function to
// construct BST from post[].
// postIndex is used to
// keep track of index in post[].
Node constructTreeUtil(int []post, Index postIndex,
int key, int min, int max, int size)
{
// Base case
if (postIndex.postindex < 0)
return null;
Node root = null;
// If current element of post[] is in range, then
// only it is part of current subtree
if (key > min && key < max)
{
// Allocate memory for root of
// this subtree and decrement *postIndex
root = new Node(key);
postIndex.postindex = postIndex.postindex - 1;
if (postIndex.postindex >= 0)
{
// All nodes which are in range
// {key..max} will go in right subtree,
// and first such node will be root of
// right subtree
root.right = constructTreeUtil(post, postIndex,
post[postIndex.postindex], key, max, size);
// Construct the subtree under root
// All nodes which are in range
// {min .. key} will go in left
// subtree, and first such node
// will be root of left subtree.
root.left = constructTreeUtil(post, postIndex,
post[postIndex.postindex],min, key, size);
}
}
return root;
}
// The main function to construct
// BST from given postorder traversal.
// This function mainly uses constructTreeUtil()
Node constructTree(int []post, int size)
{
Index index = new Index();
index.postindex = size - 1;
return constructTreeUtil(post, index,
post[index.postindex],
int.MinValue, int.MaxValue, size);
}
// A utility function to print
// inorder traversal of a Binary Tree
void printInorder(Node node)
{
if (node == null)
return;
printInorder(node.left);
Console.Write(node.data + " ");
printInorder(node.right);
}
// Driver code
public static void Main(String[] args)
{
BinaryTree tree = new BinaryTree();
int []post = new int[]{1, 7, 5, 50, 40, 10};
int size = post.Length;
Node root = tree.constructTree(post, size);
Console.WriteLine("Inorder traversal of" +
"the constructed tree:");
tree.printInorder(root);
}
}
// This code has been contributed by PrinciRaj1992
<script>
/* A O(n) program for
construction of BST from
postorder traversal */
/* A binary tree node has data,
pointer to left child and a
pointer to right child */
class Node {
constructor(data) {
this.data = data;
this.left = null;
this.right = null;
}
}
// Class containing variable
// that keeps a track of overall
// calculated postindex
class Index {
constructor() {
this.postindex = 0;
}
}
class BinaryTree {
// A recursive function to
// construct BST from post[].
// postIndex is used to
// keep track of index in post[].
constructTreeUtil(post, postIndex, key, min, max, size) {
// Base case
if (postIndex.postindex < 0) return null;
var root = null;
// If current element of post[] is in range, then
// only it is part of current subtree
if (key > min && key < max) {
// Allocate memory for root of
// this subtree and decrement *postIndex
root = new Node(key);
postIndex.postindex = postIndex.postindex - 1;
if (postIndex.postindex >= 0) {
// All nodes which are in range
// {key..max} will go in right subtree,
// and first such node will be root of
// right subtree
root.right = this.constructTreeUtil(
post,
postIndex,
post[postIndex.postindex],
key,
max,
size
);
// Construct the subtree under root
// All nodes which are in range
// {min .. key} will go in left
// subtree, and first such node
// will be root of left subtree.
root.left = this.constructTreeUtil(
post,
postIndex,
post[postIndex.postindex],
min,
key,
size
);
}
}
return root;
}
// The main function to construct
// BST from given postorder traversal.
// This function mainly uses constructTreeUtil()
constructTree(post, size) {
var index = new Index();
index.postindex = size - 1;
return this.constructTreeUtil(
post,
index,
post[index.postindex],
-2147483648,
2147483647,
size
);
}
// A utility function to print
// inorder traversal of a Binary Tree
printInorder(node) {
if (node == null) return;
this.printInorder(node.left);
document.write(node.data + " ");
this.printInorder(node.right);
}
}
// Driver code
var tree = new BinaryTree();
var post = [1, 7, 5, 50, 40, 10];
var size = post.length;
var root = tree.constructTree(post, size);
document.write("Inorder traversal of " +
"the constructed tree: <br>");
tree.printInorder(root);
</script>
OutputInorder traversal of the constructed tree:
1 5 7 10 40 50
Time Complexity: O(n)
Space Complexity: O(h), where h is the height of the BST
Note that the output to the program will always be a sorted sequence as we are printing the inorder traversal of a Binary Search Tree.
Similar Reads
Construct Full Binary Tree from given preorder and postorder traversals
Given two arrays that represent preorder and postorder traversals of a full binary tree, construct the binary tree. Full Binary Tree is a binary tree where every node has either 0 or 2 children.Examples of Full Trees. Input: pre[] = [1, 2, 4, 8, 9, 5, 3, 6, 7] , post[] = [8, 9, 4, 5, 2, 6, 7, 3, 1]O
9 min read
Construct Special Binary Tree from given Inorder traversal
Given Inorder Traversal of a Special Binary Tree in which the key of every node is greater than keys in left and right children, construct the Binary Tree and return root. Examples: Input: inorder[] = {5, 10, 40, 30, 28} Output: root of following tree 40 / \ 10 30 / \ 5 28 Input: inorder[] = {1, 5,
14 min read
Construct a BST from given postorder traversal using Stack
Given postorder traversal of a binary search tree, construct the BST.For example, If the given traversal is {1, 7, 5, 50, 40, 10}, then following tree should be constructed and root of the tree should be returned. 10 / \ 5 40 / \ \ 1 7 50 Input : 1 7 5 50 40 10 Output : Inorder traversal of the cons
9 min read
Construct Binary Tree from given Parent Array representation
Given an array that represents a tree in such a way that array indexes are values in tree nodes and array values give the parent node of that particular index (or node). The value of the root node index would always be -1 as there is no parent for root. Construct the standard linked representation o
15+ min read
Construct a Perfect Binary Tree from Preorder Traversal
Given an array pre[], representing the Preorder traversal of a Perfect Binary Tree consisting of N nodes, the task is to construct a Perfect Binary Tree from the given Preorder Traversal and return the root of the tree. Examples: Input: pre[] = {1, 2, 4, 5, 3, 6, 7}Output: 1 / \ / \ 2 3 / \ / \ / \
11 min read
Construct a special tree from given preorder traversal
Given an array pre[] that represents the Preorder traversal of a special binary tree where every node has either 0 or 2 children. One more array preLN[] is given which has only two possible values ‘L’ and ‘N’. The value ‘L’ in preLN[] indicates that the corresponding node in Binary Tree is a leaf no
12 min read
Leaf nodes from Preorder of a Binary Search Tree
Given Preorder traversal of a Binary Search Tree. Then the task is to print leaf nodes of the Binary Search Tree from the given preorder.Examples: Input: preorder[] = [4, 2, 1, 3, 6, 5]Output: [1, 3, 5]Explaination: 1, 3 and 5 are the leaf nodes as shown in the figure. Input: preorder[] = [5, 2, 10]
14 min read
Construct BST from given preorder traversal using Stack
Given preorder traversal of a binary search tree, construct the BST.For example, if the given traversal is {10, 5, 1, 7, 40, 50}, then the output should be root of following tree. 10 / \ 5 40 / \ \ 1 7 50 We have discussed O(n^2) and O(n) recursive solutions in the previous post. Following is a stac
13 min read
Construct Tree from given Inorder and Preorder traversals
Given in-order and pre-order traversals of a Binary Tree, the task is to construct the Binary Tree and return its root.Example:Input: inorder[] = [3, 1, 4, 0, 5, 2], preorder[] = [0, 1, 3, 4, 2, 5]Output: [0, 1, 2, 3, 4, 5]Explanation: The tree will look like: Table of Content[Naive Approach] Using
15+ min read
Construct a Perfect Binary Tree with given Height
Given an integer N, the task is to generate a perfect binary tree with height N such that each node has a value that is the same as its depth. Return the inorder traversal of the generated binary tree. A Perfect binary tree is a type of binary tree where every internal node has exactly two child nod
9 min read