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LC1008.java
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LC1008.java
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/*
1008. Construct Binary Search Tree from Preorder Traversal
Return the root node of a binary search tree that matches the given preorder traversal.
(Recall that a binary search tree is a binary tree where for every node,
any descendant of node.left has a value < node.val, and any descendant of node.right has a value > node.val.
Also recall that a preorder traversal displays the value of the node first, then traverses node.left, then traverses node.right.)
It's guaranteed that for the given test cases there is always possible to find a binary search tree with the given requirements.
Example 1:
Input: [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
TreeNode createNode(int val)
{
TreeNode node = new TreeNode(val);
node.left = null;
node.right = null;
return node;
}
int create_bst(int[] preorder, int n, int pos, TreeNode curr_node, int left, int right)
{
if (pos == n || preorder[pos] < left || preorder[pos] > right)
{
return pos;
}
if (preorder[pos] < curr_node.val)
{
curr_node.left = createNode(preorder[pos]);
++pos;
pos = create_bst(preorder, n, pos, curr_node.left, left, curr_node.val - 1);
}
if (pos == n || preorder[pos] < left || preorder[pos] > right)
{
return pos;
}
curr_node.right = createNode(preorder[pos]);
++pos;
pos = create_bst(preorder, n, pos, curr_node.right, curr_node.val + 1, right);
return pos;
}
public TreeNode bstFromPreorder(int[] preorder) {
int totalElem = preorder.length;
if (totalElem == 0)
{
return null;
}
TreeNode root = createNode(preorder[0]);
if (totalElem == 1)
return root;
create_bst(preorder, totalElem, 1, root, Integer.MIN_VALUE, Integer.MAX_VALUE);
return root;
}
}