DFS

Validate Binary Search Tree

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

Code:

/**
 * 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 boolean isValidBST(TreeNode root) {
        return validate(root, null, null);
    }
    
    boolean validate(TreeNode node, Integer low, Integer high) {
        // empty trees are valid
        if (node == null) return true;
        // current node's value must be between low and high.
        if ((low != null && node.val <= low) || (high != null && node.val >= high)) {
            return false;
        }
        return validate(node.right, node.val, high) && validate(node.left, low, node.val);
    }
}

• Why parameter low and high type is Integer instead of int?

  • Because we need to pass null to the function

  • If we use int, we will get compile error.

Recover Binary Search Tree

Link: https://leetcode.com/problems/recover-binary-search-tree/

Code:

/**
 * 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 void inorder(TreeNode root, List < Integer > nums) {
        if (root == null) return;
        inorder(root.left, nums);
        nums.add(root.val);
        inorder(root.right, nums);
    }

    public int[] findTwoSwapped(List < Integer > nums) {
        int n = nums.size();
        int x = -1, y = -1;
        boolean swapped_first_occurrence = false;
        for (int i = 0; i < n - 1; ++i) {
            if (nums.get(i + 1) < nums.get(i)) {
                y = nums.get(i + 1);
                if (!swapped_first_occurrence) {
                    // first swap occurrence
                    x = nums.get(i);
                    swapped_first_occurrence = true;
                } else {
                    // second swap occurrence
                    break;
                }
            }
        }
        return new int[] {
            x,
            y
        };
    }

    public void recover(TreeNode r, int count, int x, int y) {
        if (r != null) {
            if (r.val == x || r.val == y) {
                r.val = r.val == x ? y : x;
                if (--count == 0) return;
            }
            recover(r.left, count, x, y);
            recover(r.right, count, x, y);
        }
    }

    public void recoverTree(TreeNode root) {
        List < Integer > nums = new ArrayList();
        inorder(root, nums);
        int[] swapped = findTwoSwapped(nums);
        recover(root, 2, swapped[0], swapped[1]);
    }
}

The recursive inorder traversal approach:

class Solution {
    TreeNode x = null, y = null, pred = null;
    public void recoverTree(TreeNode root) {
        findTwoSwapped(root);
        swap(x, y);
    }
    
    void findTwoSwapped(TreeNode root) {
        if (root == null) return;
        findTwoSwapped(root.left);
        if (pred != null && root.val < pred.val) {
            y = root;
            if (x == null) x = pred;
            else return;
        }
        pred = root;
        findTwoSwapped(root.right);
    }
    
    void swap(TreeNode a, TreeNode b) {
        int temp = a.val;
        a.val = b.val;
        b.val = temp;
    }
}

Different DFS strategy:

DFS strategy