If the pre-order traversal of a binary search tree is 6, 2, 1, 4, 3, 7, 10, 9, 11, how to get the post-order traversal?
You are given the pre-order traversal of the tree, which is constructed by doing: output, traverse left, traverse right.
As the post-order traversal comes from a BST, you can deduce the in-order traversal (traverse left, output, traverse right) from the post-order traversal by sorting the numbers. In your example, the in-order traversal is 1, 2, 3, 4, 6, 7, 9, 10, 11.
From two traversals we can then construct the original tree. Let's use a simpler example for this:
Pre-order: 2, 1, 4, 3
In-order: 1, 2, 3, 4
The pre-order traversal gives us the root of the tree as 2. The in-order traversal tells us 1 falls into the left sub-tree and 3, 4 falls into the right sub-tree. The structure of the left sub-tree is trivial as it contains a single element. The right sub-tree's pre-order traversal is deduced by taking the order of the elements in this sub-tree from the original pre-order traversal: 4, 3. From this we know the root of the right sub-tree is 4 and from the in-order traversal (3, 4) we know that 3 falls into the left sub-tree. Our final tree looks like this:
2
/ \
1 4
/
3
With the tree structure, we can get the post-order traversal by walking the tree: traverse left, traverse right, output. For this example, the post-order traversal is 1, 3, 4, 2.
To generalise the algorithm:
The first element in the pre-order traversal is the root of the tree. Elements less than the root form the left sub-tree. Elements greater than the root form the right sub-tree.
Find the structure of the left and right sub-trees using step 1 with a pre-order traversal that consists of the elements we worked out to be in that sub-tree placed in the order they appear in the original pre-order traversal.
Traverse the resulting tree in post-order to get the post-order traversal associated with the given pre-order traversal.
Using the above algorithm, the post-order traversal associated with the pre-order traversal in the question is: 1, 3, 4, 2, 9, 11, 10, 7, 6. Getting there is left as an exercise.
Pre-order = outputting the values of a binary tree in the order of the current node, then the left subtree, then the right subtree.
Post-order = outputting the values of a binary tree in the order of the left subtree, then the right subtree, the the current node.
In a binary search tree, the values of all nodes in the left subtree are less than the value of the current node; and alike for the right subtree. Hence if you know the start of a pre-order dump of a binary search tree (i.e. its root node's value), you can easily decompose the whole dump into the root node value, the values of the left subtree's nodes, and the values of the right subtree's nodes.
To output the tree in post-order, recursion and output reordering is applied. This task is left upon the reader.
Based on Ondrej Tucny's answer. Valid for BST only
example:
20
/ \
10 30
/\ \
6 15 35
Preorder = 20 10 6 15 30 35
Post = 6 15 10 35 30 20
For a BST, In Preorder traversal; first element of array is 20. This is the root of our tree. All numbers in array which are lesser than 20 form its left subtree and greater numbers form right subtree.
//N = number of nodes in BST (size of traversal array)
int post[N] = {0};
int i =0;
void PretoPost(int pre[],int l,int r){
if(l==r){post[i++] = pre[l]; return;}
//pre[l] is root
//Divide array in lesser numbers and greater numbers and then call this function on them recursively
for(int j=l+1;j<=r;j++)
if(pre[j]>pre[l])
break;
PretoPost(a,l+1,j-1); // add left node
PretoPost(a,j,r); //add right node
//root should go in the end
post[i++] = pre[l];
return;
}
Please correct me if there is any mistake.
you are given the pre-order traversal results. then put the values to a suitable binary search tree and just follow the post-order traversal algorithm for the obtained BST.
This is the code of preorder to postorder traversal in python.
I am constructing a tree so you can find any type of traversal
def postorder(root):
if root==None:
return
postorder(root.left)
print(root.data,end=" ")
postorder(root.right)
def preordertoposorder(a,n):
root=Node(a[0])
top=Node(0)
temp=Node(0)
temp=None
stack=[]
stack.append(root)
for i in range(1,len(a)):
while len(stack)!=0 and a[i]>stack[-1].data:
temp=stack.pop()
if temp!=None:
temp.right=Node(a[i])
stack.append(temp.right)
else:
stack[-1].left=Node(a[i])
stack.append(stack[-1].left)
return root
class Node:
def __init__(self,data):
self.data=data
self.left=None
self.right=None
a=[40,30,35,80,100]
n=5
root=preordertoposorder(a,n)
postorder(root)
# print(root.data)
# print(root.left.data)
# print(root.right.data)
# print(root.left.right.data)
# print(root.right.right.data)
If you have been given preorder and you want to convert it into postorder. Then you should remember that in a BST in order always give numbers in ascending order.Thus you have both Inorder as well as the preorder to construct a tree.
preorder: 6, 2, 1, 4, 3, 7, 10, 9, 11
inorder: 1, 2, 3, 4, 6, 7, 9, 10, 11
And its postorder: 1 3 4 2 9 11 10 7 6
I know this is old but there is a better solution.
We don't have to reconstruct a BST to get the post-order from the pre-order.
Here is a simple python code that does it recursively:
import itertools
def postorder(preorder):
if not preorder:
return []
else:
root = preorder[0]
left = list(itertools.takewhile(lambda x: x < root, preorder[1:]))
right = preorder[len(left) + 1:]
return postorder(left) + postorder(right) + [root]
if __name__ == '__main__':
preorder = [20, 10, 6, 15, 30, 35]
print(postorder(preorder))
Output:
[6, 15, 10, 35, 30, 20]
Explanation:
We know that we are in pre-order. This means that the root is at the index 0 of the list of the values in the BST. And we know that the elements following the root are:
first: the elements less than the root, which belong to the left subtree of the root
second: the elements greater than the root, which belong to the right subtree of the root
We then just call recursively the function on both subtrees (which still are in pre-order) and then chain left + right + root (which is the post-order).
Here pre-order traversal of a binary search tree is given in array.
So the 1st element of pre-order array will root of BST.We will find the left part of BST and right part of BST.All the element in pre-order array is lesser than root will be left node and All the element in pre-order array is greater then root will be right node.
#include <bits/stdc++.h>
using namespace std;
int arr[1002];
int no_ans = 0;
int n = 1000;
int ans[1002] ;
int k = 0;
int find_ind(int l,int r,int x){
int index = -1;
for(int i = l;i<=r;i++){
if(x<arr[i]){
index = i;
break;
}
}
if(index == -1)return index;
for(int i =l+1;i<index;i++){
if(arr[i] > x){
no_ans = 1;
return index;
}
}
for(int i = index;i<=r;i++){
if(arr[i]<x){
no_ans = 1;
return index;
}
}
return index;
}
void postorder(int l ,int r){
if(l < 0 || r >= n || l >r ) return;
ans[k++] = arr[l];
if(l==r) return;
int index = find_ind(l+1,r,arr[l]);
if(no_ans){
return;
}
if(index!=-1){
postorder(index,r);
postorder(l+1,index-1);
}
else{
postorder(l+1,r);
}
}
int main(void){
int t;
scanf("%d",&t);
while(t--){
no_ans = 0;
int n ;
scanf("%d",&n);
for(int i = 0;i<n;i++){
cin>>arr[i];
}
postorder(0,n-1);
if(no_ans){
cout<<"NO"<<endl;
}
else{
for(int i =n-1;i>=0;i--){
cout<<ans[i]<<" ";
}
cout<<endl;
}
}
return 0;
}
As we Know preOrder follow parent, left, right series.
In order to construct tree we need to follow few basic steps-:
your question consist of series 6, 2,1,4,3,7,10,9,11
points-:
First number of series will be root(parent) i.e 6
2.Find the number which is greater than 6 so in this series 7 is first greater number in this series so right node will be starting from here and left to this number(7) is your left subtrees.
6
/ \
2 7
/ \ \
1 4 10
/ / \
3 9 11
3.same way follow the basic rule of BST i.e left,root,right
the series of post order will be L, R, N i.e. 1,3,4,2,9,11,10,7,6
Here is full code )
class Tree:
def __init__(self, data = None):
self.left = None
self.right = None
self.data = data
def add(self, data):
if self.data is None:
self.data = data
else:
if data < self.data:
if self.left is None:
self.left = Tree(data)
else:
self.left.add(data)
elif data > self.data:
if self.right is None:
self.right = Tree(data)
else:
self.right.add(data)
def inOrder(self):
if self.data:
if self.left is not None:
self.left.inOrder()
print(self.data)
if self.right is not None:
self.right.inOrder()
def postOrder(self):
if self.data:
if self.left is not None:
self.left.postOrder()
if self.right is not None:
self.right.postOrder()
print(self.data)
def preOrder(self):
if self.data:
print(self.data)
if self.left is not None:
self.left.preOrder()
if self.right is not None:
self.right.preOrder()
arr = [6, 2, 1, 4, 3, 7, 10, 9, 11]
root = Tree()
for i in range(len(arr)):
root.add(arr[i])
print(root.inOrder())
Since, it is a binary search tree, the inorder traversal will be always be the sorted elements. (left < root < right)
so, you can easily write its in-order traversal results first, which is : 1,2,3,4,6,7,9,10,11
given Pre-order : 6, 2, 1, 4, 3, 7, 10, 9, 11
In-order : left, root, right
Pre-order : root, left, right
Post-order : left, right, root
now, we got from pre-order, that root is 6.
now, using in-order and pre-order results:
Step 1:
6
/ \
/ \
/ \
/ \
{1,2,3,4} {7,9,10,11}
Step 2: next root is, using in-order traversal, 2:
6
/ \
/ \
/ \
/ \
2 {7,9,10,11}
/ \
/ \
/ \
1 {3,4}
Step 3: Similarly, next root is 4:
6
/ \
/ \
/ \
/ \
2 {7,9,10,11}
/ \
/ \
/ \
1 4
/
3
Step 4: next root is 3, but no other element is remaining to be fit in the child tree for "3". Considering next root as 7 now,
6
/ \
/ \
/ \
/ \
2 7
/ \ \
/ \ {9,10,11}
/ \
1 4
/
3
Step 5: Next root is 10 :
6
/ \
/ \
/ \
/ \
2 7
/ \ \
/ \ 10
/ \ / \
1 4 9 11
/
3
This is how, you can construct a tree, and finally find its post-order traversal, which is : 1, 3, 4, 2, 9, 11, 10, 7, 6
Related
How do we determine breadth a of binary tree.
A simple bin tree
O
/ \
O O
\
O
\
O
\
O
Breadth of above tree is 4
You could use a recursive function that returns two values for a given node: the extent of the subtree at that node towards the left (a negative number or zero), and the extent to the right (zero or positive). So for the example tree given in the question it would return -1, and 3.
To find these extends is easy when you know the extents of the left child and of the right child. And that is where the recursion kicks in, which in fact represents a depth-first traversal.
Here is how that function would look in Python:
def extents(tree):
if not tree:
# If a tree with just one node has extents 0 and 0, then "nothing" should
# have a negative extent to the right and a positive on the left,
# representing a negative breadth
return 1, -1
leftleft, leftright = extents(tree.left)
rightleft, rightright = extents(tree.right)
return min(leftleft-1, rightleft+1), max(leftright-1, rightright+1)
The breadth is simply the difference between the two extents returned by the above function, plus 1 (to count for the root node):
def breadth(tree):
leftextent, rightextent = extents(tree)
return rightextent-leftextent+1
The complete Python code with the example tree, having 6 nodes, as input:
from collections import namedtuple
Node = namedtuple('Node', ['left', 'right'])
def extents(tree):
if not tree:
return 1, -1
leftleft, leftright = extents(tree.left)
rightleft, rightright = extents(tree.right)
return min(leftleft-1, rightleft+1), max(leftright-1, rightright+1)
def breadth(tree):
left, right = extents(tree)
return right-left+1
# example tree as given in question
tree = Node(
Node(
None,
Node(None, Node(None, Node(None, None)))
),
Node(None, None)
)
print(breadth(tree)) # outputs 4
Given inorder and preorder traversals of a tree, how the tree can be re-constructed in non-recursive manner.
For example:
Re-construct the following tree
1
2 3
4 5 6 7
8 9
given
inorder traversal: 4, 2, 5, 8, 1, 6, 3, 9, 7
preorder traversal: 1, 2, 4, 5, 8, 3, 6, 7, 9
Note: There are many references to recursive implementations. For example, one may refer Construct Tree from given Inorder and Preorder traversals. However intent here is to find non-recursive implementation.
Idea is to keep tree nodes in a stack from preorder traversal, till their counterpart is not found in inorder traversal. Once a counterpart is found, all children in the left sub-tree of the node must have been already visited.
Following is non-recursive Java implementation.
public TreeNode constructTree(int[] preOrder, int[] inOrder) {
if (preOrder.length == 0) {
return null;
}
int preOrderIndex = 0;
int inOrderIndex = 0;
ArrayDeque<TreeNode> stack = new ArrayDeque<>();
TreeNode root = new TreeNode(preOrder[0]);
stack.addFirst(root);
preOrderIndex++;
while (!stack.isEmpty()) {
TreeNode top = stack.peekFirst();
if (top.val == inOrder[inOrderIndex]) {
stack.pollFirst();
inOrderIndex++;
// if all the elements in inOrder have been visted, we are done
if (inOrderIndex == inOrder.length) {
break;
}
// Check if there are still some unvisited nodes in the left
// sub-tree of the top node in the stack
if (!stack.isEmpty()
&& stack.peekFirst().val == inOrder[inOrderIndex]) {
continue;
}
// As top node in stack, still has not encontered its counterpart
// in inOrder, so next element in preOrder must be right child of
// the removed node
TreeNode node = new TreeNode(preOrder[preOrderIndex]);
preOrderIndex++;
top.right = node;
stack.addFirst(node);
} else {
// Top node in the stack has not encountered its counterpart
// in inOrder, so next element in preOrder must be left child
// of this node
TreeNode node = new TreeNode(preOrder[preOrderIndex]);
preOrderIndex++;
top.left = node;
stack.addFirst(node);
}
}
return root;
}
I was asked to implement a breadth first search for solving an eight-puzzle, representing each of its states with a vector with 9 elements, storing the tile number (or 0 for the gap) as the the data in the position.
For example [1, 3, 4, 2, 8, 0, 6, 7, 5] represents:
1 3 4
2 8 # <- Gap here
6 7 5
My pseudo-coded algorithm so far is:
startnode.configuration = start
startnode.parent = NIL
startnode.distance = 0
startnode.state = DISCOVERED
nodes_to_process = queue {}
nodes_to_process.enqueue(startnode)
while nodes_to_process is not empty
node = nodes_to_process.dequeue()
for each neighbour in NEIGHBOURS(node)
if neighbour.state == UNDISCOVERED
neighbour.distance = node.distance + 1
neighbour.parent = node
neighbour.state = DISCOVERED
The problem with this is: when I add a new neighbour how do I keep track of which nodes have been assigned as visited? Should I compare the state arrays one by one (preferaby having an ordered set of them), or is there a more refined approach, where I can skip this array-comparison step? Is the best algorithm complexity O(N * ln N) for the search?
i have a binary tree.
2
/ \
3 4
/ \ \
5 1 8
\ /
6 9
I want to change the info part of each node such that the
nodeinfo = nodeinfo + nextInorderNodeInfo
so the actual inorder traversal
5, 6, 3, 1, 2, 4, 9, 8
will change to
5+6,6+3,3+1,1+2,2+4,4+9,9+8,8+0
11, 9, 4, 3, 6, 13, 17, 8
i need to write a function that will modify the binary tree info parts of each node.
i have done the following
calling
change(root,NULL);
function definition
void change(node* n, node *k)
{
if (n)
{
if (n->left) change(n->left,n);
if (n->right) change(n,n->right);
n->info + = k->info;
}
}
in this way i am not able to modify the nodes that are right hand leaf nodes.
can someone give the correct solution..???
thanks in advance
Write a reverse in-order traversal function (as in right, this, left rather than left, this, right) (which is technically still in-order, just with a different definition, but that's besides the point).
So this function will process the nodes in this order:
8, 9, 4, 2, 1, 3, 6, 5
This function must also remember the last processed node's value (before you added to it) and simply add this value to the current node.
And here's even some code which should work:
int last = 0;
void change(node* n)
{
if (n)
{
change(n->right);
int tempLast = n->info;
n->info += last;
last = tempLast;
change(n->left);
}
}
We know the pre-order, in-order and post-order traversals. What algorithm will reconstruct the BST?
Because it is BST, in-order can be sorted from pre-order or post-order <1>. Actually, either pre-order or post-order is needed only....
<1> if you know what the comparison function is
From pre-order and in-order, to construct a binary tree
BT createBT(int* preOrder, int* inOrder, int len)
{
int i;
BT tree;
if(len <= 0)
return NULL;
tree = new BTNode;
t->data = *preOrder;
for(i = 0; i < len; i++)
if(*(inOrder + i) == *preOrder)
break;
tree->left = createBT(preOrder + 1, inOrder, i);
tree->right = createBT(preOrder + i + 1, inOrder + i + 1, len - i - 1);
return tree;
}
The rationale behind this:
In pre-order, the first node is the root. Find the root in the in-order. Then the tree can be divided into left and right. Do it recursively.
Similar for post-order and in-order.
I personally found Dante's answer a little hard to follow. I worked my way through the solution and found it to be similar to the one posted here http://geeksforgeeks.org/?p=6633
Complexity is O(N^2).
Here's another approach for building a tree using post-order traversal: http://www.technicallyidle.com/2011/02/15/build-binary-search-tree-using-post-order-traversal-trace/
Hope this helps
For reconstruction of a binary tree either preorder+inorder or postorder+inorder is needed. As already pointed out for a BST we can reconstruct using either preorder or postorder as sorting either of them will give us the inorder.
You can use the following function which is modification of the code given by #brainydexter to reconstruct the tree without using the static variable:
struct node* buildTree(char in[],char pre[], int inStrt, int inEnd,int preIndex){
// start index > end index..base condition return NULL.
if(inStrt > inEnd)
return NULL;
// build the current node with the data at pre[preIndex].
struct node *tNode = newNode(pre[preIndex]);
// if all nodes are constructed return.
if(inStrt == inEnd)
return tNode;
// Else find the index of this node in Inorder traversal
int inIndex = search(in, inStrt, inEnd, tNode->data);
// Using index in Inorder traversal, construct left and right subtress
tNode->left = buildTree(in, pre, inStrt, inIndex-1,preIndex+1);
tNode->right = buildTree(in, pre, inIndex+1, inEnd,preIndex+inIndex+1);
return tNode;
}
Here is a Ruby recursive solution
def rebuild(preorder, inorder)
root = preorder.first
root_inorder = inorder.index root
return root unless root_inorder
root.left = rebuild(preorder[1, root_inorder], inorder[0...root_inorder])
root.right = rebuild(preorder[root_inorder+1..-1], inorder[root_inorder+1..-1])
root
end
And an example
class Node
attr_reader :val
attr_accessor :left, :right
def initialize(val)
#val = val
end
def ==(node)
node.val == val
end
def inspect
"val: #{val}, left: #{left && left.val || "-"}, right: #{right && right.val || "-"}"
end
end
inorder = [4, 7, 2, 5, 1, 3, 8, 6, 9].map{|v| Node.new v }
preorder = [1, 2, 4, 7, 5, 3, 6, 8, 9].map{|v| Node.new v }
tree = rebuild(preorder, inorder)
tree
# val: 1, left: 2, right: 3
tree.left
# val: 2, left: 4, right: 5
tree.left.left
# val: 4, left: -, right: 7