Reverse Level Order Traversal

We have discussed level order traversal of a post in previous post. The idea is to print last level first, then second last level, and so on. Like Level order traversal, every level is printed from left to right.

Example Tree

Reverse Level order traversal of the above tree is “4 5 2 3 1”.
Both methods for normal level order traversal can be easily modified to do reverse level order traversal.

METHOD 1 (Recursive function to print a given level)
We can easily modify the method 1 of the normal level order traversal. In method 1, we have a method printGivenLevel() which prints a given level number. The only thing we need to change is, instead of calling printGivenLevel() from first level to last level, we call it from last level to first level.

// A recursive C program to print REVERSE level order traversal #include <stdio.h> #include <stdlib.h>   /* A binary tree node has data, pointer to left and right child */ struct node {     int data;     struct node* left;     struct node* right; };   /*Function protoypes*/ void printGivenLevel(struct node* root, int level); int height(struct node* node); struct node* newNode(int data);   /* Function to print REVERSE level order traversal a tree*/ void reverseLevelOrder(struct node* root) {     int h = height(root);     int i;     for (i=h; i>=1; i--) //THE ONLY LINE DIFFERENT FROM NORMAL LEVEL ORDER         printGivenLevel(root, i); }   /* Print nodes at a given level */ void printGivenLevel(struct node* root, int level) {     if (root == NULL)         return;     if (level == 1)         printf("%d ", root->data);     else if (level > 1)     {         printGivenLevel(root->left, level-1);         printGivenLevel(root->right, level-1);     } }   /* Compute the "height" of a tree -- the number of     nodes along the longest path from the root node     down to the farthest leaf node.*/ int height(struct node* node) {     if (node==NULL)         return 0;     else     {         /* compute the height of each subtree */         int lheight = height(node->left);         int rheight = height(node->right);           /* use the larger one */         if (lheight > rheight)             return(lheight+1);         else return(rheight+1);     } }   /* Helper function that allocates a new node with the    given data and NULL left and right pointers. */ struct node* newNode(int data) {     struct node* node = (struct node*)                         malloc(sizeof(struct node));     node->data = data;     node->left = NULL;     node->right = NULL;       return(node); }   /* 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);       printf("Level Order traversal of binary tree is \n");     reverseLevelOrder(root);       return 0; }

Output:

Level Order traversal of binary tree is
4 5 2 3 1

Time Complexity: The worst case time complexity of this method is O(n^2). For a skewed tree, printGivenLevel() takes O(n) time where n is the number of nodes in the skewed tree. So time complexity of printLevelOrder() is O(n) + O(n-1) + O(n-2) + .. + O(1) which is O(n^2).


METHOD 2 (Using Queue and Stack)
The method 2 of normal level order traversal can also be easily modified to print level order traversal in reverse order. The idea is to use a stack to get the reverse level order. If we do normal level order traversal and instead of printing a node, push the node to a stack and then print contents of stack, we get “5 4 3 2 1” for above example tree, but output should be “4 5 2 3 1”. So to get the correct sequence (left to right at every level), we process children of a node in reverse order, we first push the right subtree to stack, then left subtree.

// A C++ program to print REVERSE level order traversal using stack and queue // This approach is adopted from following link // http://tech-queries.blogspot.in/2008/12/level-order-tree-traversal-in-reverse.html #include <iostream> #include <stack> #include <queue> 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; }

Output:

Level Order traversal of binary tree is
4 5 6 7 2 3 1

Time Complexity: O(n) where n is number of nodes in the binary tree.