Create a binary tree from post order traversal and leaf node array
Last Updated :
26 Nov, 2021
Given 2 arrays, the first one containing postorder traversal sequence and the second one containing the information whether the corresponding node in the first array is a leaf node or a non-leaf node, create a binary tree and return its root and print it's inorder traversal. (There can be more than one tree possible, but you have to form only one tree)
Examples:
Input:
postorder = {40, 20, 50, 60, 30, 10}
isLeaf = {true, false, true, true, false, false}
Output: 20 40 10 50 30 60
Explanation:
Generated Binary tree
10
/ \
20 30
\ / \
40 50 60
Input:
postorder = {20, 18, 25, 100, 81, 15, 7}
isLeaf = {true, false, true, true, false, false, false}
Output: 7 18 20 15 25 81 100
Explanation:
Generated Binary tree
7
\
15
/ \
18 81
\ / \
20 25 100
Approach:
The idea is to first construct the root node of the binary tree using the last key in the post-order sequence. Then using the given boolean array, we find if the root node is an internal node or a leaf node. If the root node is an internal node, we recursively construct its right and left subtrees.
Below is the implementation of the above approach :
C++
// C++ implementation for
// the above approach
#include <bits/stdc++.h>
using namespace std;
// struct to store
// tree nodes
struct Tree {
int val;
Tree* leftchild;
Tree* rightchild;
Tree(int _val, Tree* _leftchild, Tree* _rightchild)
{
val = _val;
leftchild = _leftchild;
rightchild = _rightchild;
}
};
// Function to generate binary tree
// from given postorder traversal sequence
// and leaf or non-leaf node information.
struct Tree* createBinaryTree(int post[], bool isLeaf[], int& n)
{
// Base condition
if (n < 0){
return NULL;
}
struct Tree* root = new Tree(post[n], NULL, NULL);
bool isInternalNode = !isLeaf[n];
n--;
// If internal node
// creating left and
// right child
if (isInternalNode) {
root->rightchild = createBinaryTree(post, isLeaf, n);
root->leftchild = createBinaryTree(post, isLeaf, n);
}
return root;
}
// Function to print
// in-order traversal
// of a binary tree.
void inorder(struct Tree* root)
{
if (root == NULL){
return;
}
inorder(root->leftchild);
cout << root->val << " ";
inorder(root->rightchild);
}
// Driver code
int main()
{
int post[] = { 40, 20, 50, 60, 30, 10 };
bool isLeaf[] = { true, false, true, true, false, false };
int n = sizeof(post) / sizeof(post[0]) - 1;
struct Tree* root = createBinaryTree(post, isLeaf, n);
inorder(root);
return 0;
}
// Java implementation for
// the above approach
class GFG
{
static int n;
// to store tree nodes
static class Tree
{
int val;
Tree leftchild;
Tree rightchild;
Tree(int _val, Tree _leftchild,
Tree _rightchild)
{
val = _val;
leftchild = _leftchild;
rightchild = _rightchild;
}
};
// Function to generate binary tree
// from given postorder traversal sequence
// and leaf or non-leaf node information.
static Tree createBinaryTree(int post[],
boolean isLeaf[])
{
// Base condition
if (n < 0)
{
return null;
}
Tree root = new Tree(post[n], null, null);
boolean isInternalNode = !isLeaf[n];
n--;
// If internal node creating left and
// right child
if (isInternalNode)
{
root.rightchild = createBinaryTree(post, isLeaf);
root.leftchild = createBinaryTree(post, isLeaf);
}
return root;
}
// Function to print in-order traversal
// of a binary tree.
static void inorder(Tree root)
{
if (root == null)
{
return;
}
inorder(root.leftchild);
System.out.print(root.val + " ");
inorder(root.rightchild);
}
// Driver code
public static void main(String[] args)
{
int post[] = { 40, 20, 50, 60, 30, 10 };
boolean isLeaf[] = { true, false, true,
true, false, false };
n = post.length - 1;
Tree root = createBinaryTree(post, isLeaf);
inorder(root);
}
}
// This code is contributed by Rajput-Ji
# Python implementation of above algorithm
# Utility class to create a node
class Tree:
def __init__(self, key):
self.val = key
self.leftchild = self.rightchild = None
n = 0
# Function to generate binary tree
# from given postorder traversal sequence
# and leaf or non-leaf node information.
def createBinaryTree( post, isLeaf):
global n
# Base condition
if (n < 0):
return None
root = Tree(post[n])
isInternalNode = not isLeaf[n]
n = n - 1
# If internal node
# creating left and
# right child
if (isInternalNode):
root.rightchild = createBinaryTree(post, isLeaf)
root.leftchild = createBinaryTree(post, isLeaf)
return root
# Function to print
# in-order traversal
# of a binary tree.
def inorder( root):
if (root == None):
return
inorder(root.leftchild)
print( root.val ,end = " ")
inorder(root.rightchild)
# Driver code
post = [40, 20, 50, 60, 30, 10]
isLeaf = [True, False, True, True, False, False ]
n = len(post)-1
root = createBinaryTree(post, isLeaf)
inorder(root)
# This code is contributed by Arnab Kundu
// C# implementation of the above approach
using System;
class GFG
{
static int n;
// to store tree nodes
public class Tree
{
public int val;
public Tree leftchild;
public Tree rightchild;
public Tree(int _val, Tree _leftchild,
Tree _rightchild)
{
val = _val;
leftchild = _leftchild;
rightchild = _rightchild;
}
};
// Function to generate binary tree
// from given postorder traversal sequence
// and leaf or non-leaf node information.
static Tree createBinaryTree(int []post,
Boolean []isLeaf)
{
// Base condition
if (n < 0)
{
return null;
}
Tree root = new Tree(post[n], null, null);
Boolean isInternalNode = !isLeaf[n];
n--;
// If internal node creating left and
// right child
if (isInternalNode)
{
root.rightchild = createBinaryTree(post,
isLeaf);
root.leftchild = createBinaryTree(post,
isLeaf);
}
return root;
}
// Function to print in-order traversal
// of a binary tree.
static void inorder(Tree root)
{
if (root == null)
{
return;
}
inorder(root.leftchild);
Console.Write(root.val + " ");
inorder(root.rightchild);
}
// Driver code
public static void Main(String[] args)
{
int []post = { 40, 20, 50, 60, 30, 10 };
Boolean []isLeaf = { true, false, true,
true, false, false };
n = post.Length - 1;
Tree root = createBinaryTree(post, isLeaf);
inorder(root);
}
}
// This code is contributed by PrinciRaj1992
<script>
// JavaScript implementation of the above approach
var n;
// to store tree nodes
class Tree
{
constructor(_val, _leftchild, _rightchild)
{
this.val = _val;
this.leftchild = _leftchild;
this.rightchild = _rightchild;
}
};
// Function to generate binary tree
// from given postorder traversal sequence
// and leaf or non-leaf node information.
function createBinaryTree(post, isLeaf)
{
// Base condition
if (n < 0)
{
return null;
}
var root = new Tree(post[n], null, null);
var isInternalNode = !isLeaf[n];
n--;
// If internal node creating left and
// right child
if (isInternalNode)
{
root.rightchild = createBinaryTree(post,
isLeaf);
root.leftchild = createBinaryTree(post,
isLeaf);
}
return root;
}
// Function to print in-order traversal
// of a binary tree.
function inorder(root)
{
if (root == null)
{
return;
}
inorder(root.leftchild);
document.write(root.val + " ");
inorder(root.rightchild);
}
// Driver code
var post = [40, 20, 50, 60, 30, 10];
var isLeaf = [true, false, true,
true, false, false ];
n = post.length - 1;
var root = createBinaryTree(post, isLeaf);
inorder(root);
</script>
Output: 20 40 10 50 30 60
Time Complexity: O(N).
Auxiliary Space: O(N).
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