Pre-order to post-order traversal

Question

If the pre-order traversal of a binary search tree is 6, 2, 1, 4, 3, 7, 10, 9, 11, how to get the post-order traversal?

Answer 1

Since, it is a binary search tree, the inorder traversal will be always be the sorted elements. (left < root < right)

so, you can easily write its in-order traversal results first, which is : 1,2,3,4,6,7,9,10,11

given Pre-order : 6, 2, 1, 4, 3, 7, 10, 9, 11

In-order : left, root, right Pre-order : root, left, right Post-order : left, right, root

now, we got from pre-order, that root is 6.

now, using in-order and pre-order results: Step 1:

             6
            / \
           /   \
          /     \
         /       \
   {1,2,3,4}  {7,9,10,11}

Step 2: next root is, using in-order traversal, 2:

             6
            / \
           /   \
          /     \
         /       \
        2  {7,9,10,11}
       / \
      /   \
     /     \
    1    {3,4}

Step 3: Similarly, next root is 4:

             6
            / \
           /   \
          /     \
         /       \
        2  {7,9,10,11}
       / \
      /   \
     /     \
    1       4
           /
          3

Step 4: next root is 3, but no other element is remaining to be fit in the child tree for "3". Considering next root as 7 now,

             6
            / \
           /   \
          /     \
         /       \
        2         7
       / \         \
      /   \       {9,10,11}
     /     \
    1       4
           /
          3

Step 5: Next root is 10 :

             6
            / \
           /   \
          /     \
         /       \
        2         7
       / \         \
      /   \         10
     /     \       /  \
    1       4     9   11
           /
          3

This is how, you can construct a tree, and finally find its post-order traversal, which is : 1, 3, 4, 2, 9, 11, 10, 7, 6