You have been given a Binary Tree of 'N' nodes, where the nodes have integer values. Your task is to find the Pre-Order traversal of the given binary tree.
For example :
For the given binary tree:
The Preorder traversal will be [1, 3, 5, 2, 4, 7, 6].
The first line contains an integer 'T' which denotes the number of test cases.
The first 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 Pre-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 Pre-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
1 2 3 6
1 2 3
In test case 1, the given binary tree is shown below:
Preorder traversal of given tree = [1, 2, 3, 6]
In test case 2, the given binary tree is shown below:
Preorder traversal of given tree = [1, 2, 3]
2
1 -1 -1
1 2 4 5 3 -1 -1 -1 -1 -1 -1
1
1 2 5 3 4
In test case 1, there is only one node, so Pre-Order traversal will be only [1].
In test case 2, the given binary tree is shown below:
Preorder traversal of given tree = [1, 2, 5, 3, 4]
Think of a recursive solution.
As we can see, only after processing any node, the left subtree is processed, followed by the right subtree. 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 ).
All tags
Sort by
Interview problems
Preorder Traversal || Recursion || Java || O(N)
public static List < Integer > getPreOrderTraversal(TreeNode root) {
List<Integer> treelist = new ArrayList<>();
preOrderTraversal(root, treelist);
return treelist;
}
public static void preOrderTraversal(TreeNode root, List<Integer> treelist){
if(root == null){
return;
}
treelist.add(root.data);
preOrderTraversal(root.left, treelist);
preOrderTraversal(root.right, treelist);
}
Interview problems
Easy for C++
Time complexity O(n)
space complexity O(n)
Interview problems
Preorder-Iterative : Note make sure u comment this import statement
//// import javax.swing.tree.TreeNode;
if u find this comment it to work !
public class Solution {
public static List < Integer > getPreOrderTraversal(TreeNode root) {
// Write your code here.
List<Integer> ans = new ArrayList<>();
TreeNode current = root;
Stack<TreeNode> st = new Stack<>();
while(current!=null || !st.isEmpty())
{
while(current!=null)
{
ans.add(current.data);
st.push(current);
current = current.left;
}
current = st.peek();
current = current.right;
st.pop();
}
return ans;
}
}
Interview problems
C++ Iterative + recursive
Iterative
vector<int> getPreOrderTraversal(TreeNode *root)
{
// Write your code here.
vector<int> res;
stack<TreeNode *> st;
if(root == NULL) return res;
st.push(root);
while(!st.empty()){
TreeNode* temp = st.top();
st.pop();
if(temp->right != NULL) st.push(temp->right);
if(temp->left != NULL) st.push(temp->left);
res.push_back(temp->data);
}
return res;
}
RECURSIVE:
class Solution {
public:
void preorder(TreeNode* root, vector<int> &res){
if(root == nullptr) return;
res.push_back(root->val);
preorder(root->left, res);
preorder(root->right, res);
}
vector<int> getPreOrderTraversal(TreeNode *root) {
vector<int> res;
preorder(root, res);
return res;
}
};
Interview problems
Easy Iterative Solution C++
vector<int> getPreOrderTraversal(TreeNode *root)
{
vector<int> ans;
if(root==NULL)return ans;
stack<TreeNode*> s;
s.push(root);
while(!s.empty()){
root=s.top();
ans.push_back(root->data);
s.pop();
if(root->right!=NULL)s.push(root->right);
if(root->left!=NULL)s.push(root->left);
}
return ans;
}
Interview problems
C++ SOLUTION [Recursive]
void preOrder(TreeNode *root, vector<int> &ans)
{
if(root == NULL)
return;
ans.push_back(root->data);
preOrder(root->left, ans);
preOrder(root->right, ans);
}
vector<int> getPreOrderTraversal(TreeNode *root)
{
vector<int> ans;
preOrder(root,ans);
return ans;
}Interview problems
simple code in c++
#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) {}
};
*/
void preorder(TreeNode *root,vector<int>&ans){
if(root){
ans.push_back(root->data);
preorder(root->left,ans);
preorder(root->right,ans);
}
}
vector<int> getPreOrderTraversal(TreeNode *root)
{
vector<int> ans;
preorder(root,ans);
return ans;
// Write your code here.
}
Interview problems
C++ solution using recursion
void solve(TreeNode *root,vector<int>&answer)
{
if(root==NULL)
{
return;
}
answer.push_back(root->data);
solve(root->left,answer);
solve(root->right,answer);
}
vector<int> getPreOrderTraversal(TreeNode *root)
{
vector<int>answer;
solve(root,answer);
return answer;
}
Interview problems
CPP morris preorder traversal tc-O(n) & sc-O(1)
#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> getPreOrderTraversal(TreeNode *root)
{
vector<int>v;
TreeNode*cur=root;
while(cur!=NULL){
if(cur->left==NULL){
v.push_back(cur->data);
cur=cur->right;
}
else{
TreeNode *temp=cur->left;
while(temp->right!=NULL && temp->right!=cur){
temp=temp->right;
}
if(temp->right==NULL){
v.push_back(cur->data);
temp->right=cur;
cur=cur->left;
}else{
temp->right=NULL;
cur=cur->right;
}
}
}
return v;
}
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> pin;
void preorder(TreeNode* root){
if(root==NULL) return ;
pin.push_back(root->data);
preorder(root->left);
preorder(root->right);
}
vector<int> getPreOrderTraversal(TreeNode *root)
{
// Write your code here.
pin.clear();
preorder(root);
return pin;
}