You have been given a Binary Tree of 'N' nodes, where the nodes have integer values. Your task is to find the Post-Order traversal of the given binary tree.
For example :
For the given binary tree:
The Postorder traversal will be [5, 2, 3, 7, 6, 4, 1].
The first line contains an integer 'T' which denotes the number of test cases.
The first and only line of each test case contains elements of the tree in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place.
The input for the tree is depicted in the below image:
1 3 8 5 2 7 -1 -1 -1 -1 -1 -1 -1
Explanation :
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 3
Right child of 1 = 8
Level 3 :
Left child of 3 = 5
Right child of 3 = 2
Left child of 8 =7
Right child of 8 = null (-1)
Level 4 :
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 2 = null (-1)
Right child of 2 = null (-1)
Left child of 7 = null (-1)
Right child of 7 = null (-1)
1
3 8
5 2 7 -1
-1 -1 -1 -1 -1 -1
1. The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
2. The input ends when all nodes at the last level are null(-1).
3. The above format was just to provide clarity on how the input is formed for a given tree. The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:
1 3 8 5 2 7 -1 -1 -1 -1 -1 -1 -1
For each test case, return a vector containing the Post-Order traversal of a given binary tree.
The first and only line of output of each test case prints 'N' single space-separated integers denoting the node's values in Post-Order traversal.
You don't need to print anything, it has already been taken care of. Just implement the given function.
1 <= T <= 10
0 <= N <= 3000
0 <= data <= 10^9
Where 'data' denotes the node value of the binary tree nodes.
Time limit: 1 sec
2
1 2 3 -1 -1 -1 6 -1 -1
1 2 3 -1 -1 -1 -1
2 6 3 1
2 3 1
In test case 1, the given binary tree is shown below:
Postorder traversal of given tree = [2, 6, 3, 1]
In test case 2, the given binary tree is shown below:
Postorder traversal of given tree = [2, 3, 1]
2
1 -1 -1
1 2 4 5 3 -1 -1 -1 -1 -1 -1
1
5 3 2 4 1
In test case 1, there is only one node, so Post-Order traversal will be only [1].
In test case 2, the given binary tree is shown below:
Postorder traversal of given tree = [5, 3, 2, 4, 1]
Think of a recursive solution.
As we can see, before processing any node, the left subtree is processed first, followed by the right subtree, and the node is processed at last. These operations can be defined recursively for each node. The recursive implementation is referred to as a Depth-first search (DFS), as the search tree is deepened as much as possible on each child before going to the next sibling.
The steps are as follows :
O( N ), where ‘N’ is the total number of nodes in the given binary tree.
The recurrence relation is : T( N ) = 2 * T( N / 2 ) + O( 1 ). Apply Master theorem case : c < logba where a = 2 , b = 2 , c = 0. For master theorem http://en.wikipedia.org/wiki/Master_theorem.
Or in simple words, we visit each node of the tree exactly once.
Hence the time complexity is O( N ).
O( N ), where ‘N’ is the total number of nodes in the given binary tree.
The space required is proportional to the tree’s height, which can be equal to the total number of nodes in the tree in the worst case for skewed trees.
Hence the space complexity is O( N ).
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Interview problems
Postorder Traversal || Recursion || O(N) || Java
public static List < Integer > getPostOrderTraversal(TreeNode root) {
List<Integer> treelist = new ArrayList<>();
postOrderTraversal(root, treelist);
return treelist;
}
public static void postOrderTraversal(TreeNode root, List<Integer> treelist){
if(root == null){
return;
}
postOrderTraversal(root.left, treelist);
postOrderTraversal(root.right, treelist);
treelist.add(root.data);
}
Interview problems
SOL USING RECURSION
import java.util.*;
import java.io.*;
public class Solution {
public static List < Integer > getPostOrderTraversal(TreeNode root) {
// Create answer array to store traversal.
List < Integer > answer = new ArrayList < Integer > ();
// Call postOrderHelper function and store preOrder traversal in answer array.
postOrderHelper(root, answer);
// Return answer.
return answer;
}
// Helper function.
public static void postOrderHelper(TreeNode node, List < Integer > answer) {
// Base case.
if (node == null) {
return;
}
// Visit left subtree.
postOrderHelper(node.left, answer);
// Visit right subtree.
postOrderHelper(node.right, answer);
// Add data of node to answer.
answer.add(node.data);
}
}Interview problems
Iterative C++ Solution || Striver
#include <bits/stdc++.h>
/*
Following is Binary Tree Node structure:
class TreeNode
{
public:
int data;
TreeNode *left, *right;
TreeNode() : data(0), left(NULL), right(NULL) {}
TreeNode(int x) : data(x), left(NULL), right(NULL) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : data(x), left(left), right(right) {}
};
*/
vector<int> getPostOrderTraversal(TreeNode *root)
{
// Write your code here.
TreeNode* curr = root;
vector<int> v;
if(!root){return v;}
stack<TreeNode*> st;
while(curr!=nullptr || !st.empty()){
if(curr){
st.push(curr);
curr=curr->left;
}
else{
auto temp = st.top()->right;
if(temp==nullptr){
temp=st.top();
st.pop();
v.push_back(temp->data);
while(!st.empty() && temp == st.top()->right){
temp=st.top();
st.pop();
v.push_back(temp->data);
}
}
else{
curr=temp;
}
}
}
return v;
}Interview problems
Easy Iterative C++ Sol
vector<int> getPostOrderTraversal(TreeNode *root)
{
vector<int>ans;
if(root==NULL) return ans;
stack<TreeNode *>st;
st.push(root);
while(!st.empty()){
TreeNode * front=st.top();
st.pop();
if(front->left) st.push(front->left);
if(front->right) st.push(front->right);
ans.push_back(front->data);
}
reverse(ans.begin(),ans.end());
return ans;
}
Interview problems
Java Solution
public class Solution {
public static void postOrderTraversal(TreeNode root, List<Integer> list) {
if (root == null) return;
postOrderTraversal(root.left, list);
postOrderTraversal(root.right, list);
list.add(root.data);
}
public static List < Integer > getPostOrderTraversal(TreeNode root) {
// Write your code here.
List<Integer> list = new ArrayList<>();
postOrderTraversal(root, list);
return list;
}
}Interview problems
C++ SOLUTION [Iterative]
void solve(TreeNode *root, vector<int> &ans)
{
if(root == NULL)
return;
solve(root->left,ans);
solve(root->right,ans);
ans.push_back(root->data);
}
vector<int> getPostOrderTraversal(TreeNode *root)
{
vector<int> ans;
solve(root,ans);
return ans;
}Interview problems
C++ solution using recursion
void solve(TreeNode *root,vector<int>&answer)
{
if(root==NULL)
{
return ;
}
solve(root->left,answer);
solve(root->right,answer);
answer.push_back(root->data);
}
vector<int> getPostOrderTraversal(TreeNode *root)
{
vector<int>answer;
solve(root,answer);
return answer;
}
Interview problems
C++ || EASY TO UNDERSTAND || RECURSION
#include <bits/stdc++.h>
/*
Following is Binary Tree Node structure:
class TreeNode
{
public:
int data;
TreeNode *left, *right;
TreeNode() : data(0), left(NULL), right(NULL) {}
TreeNode(int x) : data(x), left(NULL), right(NULL) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : data(x), left(left), right(right) {}
};
*/
vector<int> post;
void postorder(TreeNode* root){
if(root==NULL) return;
postorder(root->left);
postorder(root->right);
post.push_back(root->data);
}
vector<int> getPostOrderTraversal(TreeNode *root)
{
// Write your code here.
post.clear();
postorder(root);
return post;
}Interview problems
Best Iterative solution
vector<int> getPostOrderTraversal(TreeNode *root)
{
vector<int> res;
// Write your code here.
stack<pair<TreeNode*, int>> sack;
if(root != NULL) sack.push({root,0});
while(!sack.empty()){
TreeNode *k = sack.top().first;
int state = sack.top().second;
sack.pop();
if(state == 3) continue;
sack.push({k, state+1});
if(state == 0 && k->left != NULL){
sack.push({k->left,0});
}
else if(state == 1 && k->right!= NULL){
sack.push({k->right,0});
}
else if(state == 2){
res.push_back(k->data);
}
}
return res;
} Every node has 4 states →
for inorder → (0 → check left 1→node. 2→right 3→ done)
for preorder → (0 →node 1→.check left 2→check right 3→ done)
postorder → (0 →.check left 1→.check right 2→ node 3→ done)
Interview problems
PYTHON SOL
from os import *
from sys import *
from collections import *
from math import *
'''
Following is the Binary Tree node structure:
class TreeNode:
def __init__(self, data=0, left=None, right=None):
self.data = data
self.left = left
self.right = right
'''
def postorder(root,ans):
if root:
postorder(root.left,ans)
postorder(root.right,ans)
ans.append(root.data)
def getPostOrderTraversal(root):
# Write your code here.
ans=[]
postorder(root,ans)
return ans
