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BinaryTree_ReverseLevelOrder.cpp
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BinaryTree_ReverseLevelOrder.cpp
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#include <bits/stdc++.h>
using namespace std;
/* A binary tree node has data, pointer to left and right children */
struct node
{
int data;
struct node* left;
struct node* right;
};
/* Given a binary tree, print its nodes in reverse level order */
void reverseLevelOrder(node* root)
{
stack <node *> S;
queue <node *> Q;
Q.push(root);
// Do something like normal level order traversal order. Following are the
// differences with normal level order traversal
// 1) Instead of printing a node, we push the node to stack
// 2) Right subtree is visited before left subtree
while (Q.empty() == false)
{
/* Dequeue node and make it root */
root = Q.front();
Q.pop();
S.push(root);
/* Enqueue right child */
if (root->right)
Q.push(root->right); // NOTE: RIGHT CHILD IS ENQUEUED BEFORE LEFT
/* Enqueue left child */
if (root->left)
Q.push(root->left);
}
// Now pop all items from stack one by one and print them
while (S.empty() == false)
{
root = S.top();
cout << root->data << " ";
S.pop();
}
}
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
node* newNode(int data)
{
node* temp = new node;
temp->data = data;
temp->left = NULL;
temp->right = NULL;
return (temp);
}
/* Driver program to test above functions*/
int main()
{
struct node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->left = newNode(6);
root->right->right = newNode(7);
cout << "Level Order traversal of binary tree is \n";
reverseLevelOrder(root);
return 0;
}