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Program to perform level order traversal of binary tree in C++
Suppose we have a binary tree. We have to traverse this tree using level order traversal fashion. So if the tree is like
The traversal sequence will be like: [1,2,3,5,4]
To solve this, we will follow these steps −
define queue que to store nodes
insert root into the que.
while que is not empty, do
item := item present at front position of queue
print the value of item
if left of the item is not null, then insert left of item into que
if right of the item is not null, then insert right of item into que
delete front element from que
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<int> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
class TreeNode{
public:
int val;
TreeNode *left, *right;
TreeNode(int data){
val = data;
left = right = NULL;
}
};
class Solution {
public:
vector<int> solve(TreeNode* root) {
if(!root)
return {};
vector <int> ret;
queue <TreeNode*> q;
q.push(root);
while(!q.empty()){
TreeNode* node = q.front();
q.pop();
ret.push_back(node->val);
if(node->left){
q.push(node->left);
}
if(node->right){
q.push(node->right);
}
}
return ret;
}
};
main(){
TreeNode *root = new TreeNode(1);
root->left = new TreeNode(2);
root->right = new TreeNode(3);
root->left->right = new TreeNode(5);
root->right->right = new TreeNode(4);
Solution ob;
print_vector(ob.solve(root));
}
Input
TreeNode *root = new TreeNode(1); root->left = new TreeNode(2); root->right = new TreeNode(3); root->left->right = new TreeNode(5); root->right->right = new TreeNode(4);
Output
[1, 2, 3, 5, 4, ]
Updated on: 10-Oct-2020
138 Views
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