98. Validate Binary Search Tree (Medium)

https://leetcode.com/problems/validate-binary-search-tree/

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

  • The left subtree of a node contains only nodes with keys less than the node's key.
  • The right subtree of a node contains only nodes with keys greater than the node's key.
  • Both the left and right subtrees must also be binary search trees.

Example 1:

Input:
    2
   / \
  1   3
Output: true

Example 2:

    5
   / \
  1   4
     / \
    3   6
Output: false
Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value
             is 5 but its right child's value is 4.

Solutions

class Solution {

    // Actually, the best solution is to traverse the tree and return the visiting sequence.
    // Then sort out if the outcome is sorted by ascending order or not. If not well ordered,
    // this tree must not be the valid Binary Sort Tree.

    public boolean isValidBST(TreeNode root) {
        List<Integer> track = new ArrayList<>();

        traverse(root, track);

        for (int i = 1; i < track.size(); i++) {
            if (track.get(i - 1) >= track.get(i)) {
                return false;
            }
        }

        return true;
    }

    private void traverse(TreeNode root, List<Integer> track) {
        if (root == null) {
            return;
        }

        traverse(root.left, track);

        track.add(root.val);

        traverse(root.right, track);
    }
}

Incorrect Solutions

class Solution2 {

    // This solution is trivial, nasty and incorrect.

    // You can look into this test case: [3,1,5,0,2,4,6,null,null,null,3] and will
    // find out what's going wrong here.

    public boolean isValidBST(TreeNode root) {
        if (root == null) {
            return true;
        }

        return check(root, "", root.val);
    }

    private boolean check(TreeNode root, String from, int limit) {
        if (root == null) {
            return true;
        }

        // 1 Make sure the value order is left < root < right = is also not permitted
        if (root.left != null && root.left.val >= root.val) {
            return false;
        }

        if (root.right != null && root.right.val <= root.val) {
            return false;
        }

        // 2.1 If this branch is the left one of parent, this.right.val < parent.val;
        if (from.equals("left") && root.right != null && root.right.val >= limit) {
            return false;
        }

        // 2.2 If this branch is the right one of parent, this.left.val > parent.val;
        if (from.equals("right") && root.left != null && root.left.val <= limit) {
            return false;
        }

        // If left branch is invalid, not need to validate right branch, return immediately.
        if (!check(root.left, "left", root.val)) {
            return false;
        }

        if (!check(root.right, "right", root.val)) {
            return false;
        }

        return true;
    }
}

References

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