## Saturday, March 5, 2016

### Construct a Binary Search Tree from given postorder - GeeksforGeeks

Construct a Binary Search Tree from given postorder - GeeksforGeeks
Related: Construct BST from given preorder traversal
Given postorder traversal of a binary search tree, construct the BST.
For example, if the given traversal is {1, 7, 5, 50, 40, 10}, then following tree should be constructed and root of the tree should be returned.

```     10
/   \
5     40
/  \      \
1    7      50```
Method 2 ( O(n) time complexity )
The trick is to set a range {min .. max} for every node. Initialize the range as {INT_MIN .. INT_MAX}. The last node will definitely be in range, so create root node. To construct the left subtree, set the range as {INT_MIN …root->data}. If a values is in the range {INT_MIN .. root->data}, the values is part part of left subtree. To construct the right subtree, set the range as {root->data .. INT_MAX}.
`// A recursive function to construct BST from post[]. `
`// postIndex is used to keep track of index in post[].`
`struct` `node* constructTreeUtil(``int` `post[], ``int``* postIndex,`
`                         ``int` `key, ``int` `min, ``int` `max, ``int` `size)`
`{`
`    ``// Base case`
`    ``if` `(*postIndex < 0)`
`        ``return` `NULL;`
`    ``struct` `node* root = NULL;`
`    ``// If current element of post[] is in range, then`
`    ``// only it is part of current subtree`
`    ``if` `(key > min && key < max)`
`    ``{`
`        ``// Allocate memory for root of this subtree and decrement`
`        ``// *postIndex`
`        ``root = newNode(key);`
`        ``*postIndex = *postIndex - 1;`
`        ``if` `(*postIndex > 0)`
`        ``{`
`          ``// All nodes which are in range {key..max} will go in right`
`          ``// subtree, and first such node will be root of right subtree.`
`          ``root->right = constructTreeUtil(post, postIndex, post[*postIndex],`
`                                          ``key, max, size );`
`          ``// Contruct the subtree under root`
`          ``// All nodes which are in range {min .. key} will go in left`
`          ``// subtree, and first such node will be root of left subtree.`
`          ``root->left = constructTreeUtil(post, postIndex, post[*postIndex],`
`                                         ``min, key, size );`
`        ``}`
`    ``}`
`    ``return` `root;`
`}`
`// The main function to construct BST from given postorder`
`// traversal. This function mainly uses constructTreeUtil()`
`struct` `node *constructTree (``int` `post[], ``int` `size)`
`{`
`    ``int` `postIndex = size-1;`
`    ``return` `constructTreeUtil(post, &postIndex, post[postIndex],`
`                             ``INT_MIN, INT_MAX, size);`
`}`

Method 1 ( O(n^2) time complexity )
The last element of postorder traversal is always root. We first construct the root. Then we find the index of last element which is smaller than root. Let the index be ‘i’. The values between 0 and ‘i’ are part of left subtree, and the values between ‘i+1′ and ‘n-2′ are part of right subtree. Divide given post[] at index “i” and recur for left and right sub-trees.
For example in {1, 7, 5, 40, 50, 10}, 10 is the last element, so we make it root. Now we look for the last element smaller than 10, we find 5. So we know the structure of BST is as following.
```             10
/    \
/      \
{1, 7, 5}       {50, 40}```
We recursively follow above steps for subarrays {1, 7, 5} and {40, 50}, and get the complete tree.
http://stackoverflow.com/questions/21173541/construct-binary-search-tree-from-post-order-traversal-in-java
``````public static Node buildBinarytreefromPostOrder(int[] post, int start, int end)
{
if (end < start)
return null;

Node root = new Node(post[end]);

if (end == start)
return root;

int i;
for (i = end; i >= start; i--)
if (post[i] < root.data)
break;

root.left = buildBinarytreefromPostOrder(post, start, i);
root.right = buildBinarytreefromPostOrder(post, i + 1, end - 1);

return root;
}``````
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