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Postorder traversal of a binary tree
struct NODE
{
struct NODE *left;
int value; /* can take any data type */
struct NODE *right;
};
postorder(struct NODE *curr)
if(curr->left != NULL) postorder(curr->left);
if(curr->right != NULL) postorder(curr->right);
printf("%d", curr->value);
}
In a preorder traversal, first (pre) the root is visited and then the left & right sub- trees are traversed. In a postorder traversal, first the left sub-tree is visited, followed by right sub-tree which is then followed through root. In inorder traversal, first the left sub- tree is visited, followed by root, followed by right sub-tree.
One of the main problems with the linear queue is the lack of appropriate utilization of space. Assume that the queue can store 100 elements & the complete queue is full. Thus, it
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Example 1: Following are Simple sequence of statements Statement 1; Statement 2; ... ... Statement k; The entire time can be found out through adding the times for
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Post-order Traversal This can be done both iteratively and recursively. The iterative solution would need a change of the in-order traversal algorithm.
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