Construct Binary Search Tree from Preorder Traversal

Amazon

2020-05-24

1008. Construct Binary Search Tree from Preorder Traversal

Question:

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]

Constraints:

1 <= preorder.length <= 100
1 <= preorder[i] <= 10^8
The values of preorder are distinct.

Solution:

Becuase it’s a binary search tree, all the value larger than the first node are on the right. Same for the value smaller than the first node. Thus, set a threshold for the maximum value and use recursion to assign the value to left or right.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int index = 0;
    public TreeNode bstFromPreorder(int[] preorder) {
        if (preorder == null || preorder.length == 0) return new TreeNode();
        return bstHelper(preorder, Integer.MAX_VALUE);
    }
    
    public TreeNode bstHelper(int[] preorder, int max) {
        if (index == preorder.length || preorder[index] > max) return null;
        TreeNode curr = new TreeNode(preorder[index++]);
        curr.left = bstHelper(preorder, curr.val);
        curr.right = bstHelper(preorder, max);
        return curr;
    }
}

In addition, using the same idea, we could use two pointers to record the lower bound and upper bound of the traversal.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode bstFromPreorder(int[] preorder) {
        if (preorder == null || preorder.length == 0) return new TreeNode();
        return bstHelper(preorder, 0, preorder.length - 1);
    }
    
    public TreeNode bstHelper(int[] preorder, int start, int end) {
        if (start > end) return null;
        int currval = preorder[start];
        TreeNode curr = new TreeNode(currval);
        int i;
        for (i = start; i <= end; i++) {
            if (preorder[i] > currval) {
                break;
            }
        }
        curr.left = bstHelper(preorder, start + 1, i - 1);
        curr.right = bstHelper(preorder, i, end);
        return curr;
    }
}