L24. Right/Left View of Binary Tree | C++ | Java

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Right/Left View of Binary Tree | C++ | Java

[takeUforward]

Hi, hope you are well.


Find DSA, LLD, OOPs, Core Subjects, 1000+ Premium Questions company wise, Aptitude, SQL and many other time saving features under one affordable subscription. Have a hassle free one stop solution for up-skilling and preparing yourself.

[thunder-storm-power]

I feel so lucky to have a teacher like you, thanks a lottt!!!

[uRamPlus]

self notes:
🚀 for every level, the first node (on the right side) will be our right side view
🚀 if the level of the tree == my vector's size, I need to push it into my vector
🚀 if at any point we reach a null node, we just need to return (base case)

[kuldeepkushwah]

Thank you striver for all your effort, your content is worth more than any paid courses.
This Chanel is full of treasure.

[anmolswarnkar]

Amazing explanation! I like the fact that you go step by step (like a compiler would) over the example. Keep posting these videos!

[animeshbarole]

if(level==ds.size()) ans.push_back(node->val) is just Mind blowing technique Take a bow Striver .

[takeUforward]

Please likeeee, shareeee and :) Also follow me at Insta: Striver_79

[krishnasudan]

My iterative approach.. thanks Bhaiya for everything.. Such a gem in coding community

vector<int> rightView(Node *root){
vector<int> res;
if(root==NULL) return res;
queue<Node*> q;
q.push(root);
while(!q.empty()){
Node *temp = q.front();
res.push_back(temp->data);
int size = q.size();
for(int i=0;i<size;i++){
Node *curr = q.front();
q.pop();
if(curr->right!=NULL) q.push(curr->right);
if(curr->left!=NULL) q.push(curr->left);
}
}
return res;
}

[Dontpushyour_luck]

so clever technique of using recursion and size of data structure to check if it is the first node that we came to in this level. no wonder you are candidate master on codeforces!

[VishalGupta-xwrp]

What i came up was that (before watching video)

For(Right View)
Use Vertical order only
This time instead of left - 1, we will increase both and Mark each node *levels* not their *vertical* left + 1, right + 1

Now simply as we go we will store the latest level by m[level] = node->data.

Because of map property, it will itself ensure that all the latest one will be there so overwrites the previous

But what a technique and Amazing approach.... Man when will i begin to think like that ♥️🔥

[shubh]

your explanation techniques are phenomenal, clear concepts and concise code. Loving the way you code.

[divyansh]

Easy solution for left view level order traversal:


#include <bits/stdc++.h>
using namespace std;

vector<int> getLeftView(TreeNode<int> *root)
{
if (root == NULL)
return {};
vector<int> ans;
queue<TreeNode<int> *> q;
q.push(root);

while (!q.empty())
{
int sz = q.size();
for (int i = 0; i < sz; i++)
{
auto front = q.front();
q.pop();
if (i == 0)
ans.push_back(front->data);
if (front->left)
q.push(front->left);
if (front->right)
q.push(front->right);
}
}
return ans;
}

[PalakMittal]

In case anyone want iterative method 😅
class Solution
{
public:
//Function to return list containing elements of right view of binary tree.
vector<int> rightView(Node *root)
{
// Your Code here
vector<int>res;
if(!root)
return res;
queue<Node*>q;

q.push(root);
while(!q.empty()){
int n=q.size();
for(int i=0;i<n;i++){


if(i==0){

}


Node* node=q.front();
q.pop();


if(node->right)
q.push(node->right);


if(node->left)
q.push(node->left);
}

}
return res;
}
};
PS: Recursive one was just mind blowing 😍😍

[mulate]

My iterative approach which was before I watched your explanation:
//just push to ans vector when we are at last node in the queue.

class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
vector<int> ans;
if(root==nullptr) return {};
queue<TreeNode*> q;
q.push(root);
while(!q.empty()){
int size = q.size();
for(int i=0;i<size;i++){
TreeNode* front = q.front();
if(i==size-1){
ans.push_back(front->val);
}
q.pop();
if(front->left!=nullptr) q.push(front->left);
if(front->right!=nullptr) q.push(front->right);
}
}
return ans;
}
};

[falgunitagadkar]

Your approach to solve problems is just commendable👏👏
Thanks for providing so valuable content for free!
Again...the same thing...Hats off to you Striver ✌

[samarthsingh]

I first watched the video of apna college youtube channel where they discussed the iterative method using level order traversal. This code here in this video is very short and also very well explained by striver. Thanks TUF for such amazing content.

[rohan]

For Left Side View, call the recursive function with the left node and then with the right node .
f(node->left, level+1);
f(node->right, level+1);

[rajnishism]

For Left Side View, call the recurssive function with left node and then with the right node .
f(node->left, level+1);
f(node->right, level+1);

[yogeshjoshi]

I saw the video of top view, then I solved botom view on my own then I just changed line-1 to line+1 in that bottom view solution and I got this answer right. just a single change thank you striver

[maradanikhil]

Here is the level order traversal for right side view:
class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
queue<TreeNode*>q;
q.push(root);
vector<int>ans;
if(root==NULL)
return ans;
while(!q.empty()){
int size=q.size();
int a=0;
for(int i=0;i<size;i++){
TreeNode* node=q.front();
q.pop();
a=node->val;
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
}
ans.push_back(a);
}return ans;
}
};