LeetCode-099-Restore Binary Search Tree

Restore binary search tree

Title description: Give you the root node root of the binary search tree. Two nodes in the tree have been swapped by mistake. Please restore this tree without changing its structure.

Advanced: The solution using O(n) space complexity is easy to implement. Can you think of a solution that uses only constant space?

Please refer to LeetCode official website for example description.

Source: LeetCode
Link: https://leetcode-cn.com/problems/recover-binary-search-tree/
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Solution one: recursive method

• First, get all the nodes allNodes of the binary search tree through in-order traversal. Under normal circumstances, if no nodes are incorrectly exchanged, all nodes of allNodes should be arranged in ascending order, so find out the 2 nodes that are exchanged;
• Traverse allNodes from back to front, and find the first node with a smaller value than the previous value, which is the first node high in error;
• From high-1 traverse allNodes forward and find the first node low with a value smaller than the high node. The node at low+1 is the wrong second node;
• The tree can be restored by swapping the values of the low and high 2 nodes.

import java.util.ArrayList;
import java.util.List;

public class LeetCode_099 {
    public static void recoverTree(TreeNode root) {
        List<TreeNode> allNodes = inOrder(root);
        int high = -1, low = -1;
        for (int i = allNodes.size() - 1; i >= 1; i--) {
            // 找到上面的要交换的节点
            if (allNodes.get(i).val < allNodes.get(i - 1).val) {
                high = i;
                break;
            }

        }

        // 找到下面要交换的节点
        for (low = high - 1; low >= 0; low--) {
            if (allNodes.get(low).val < allNodes.get(high).val) {
                break;
            }
        }
        low++;
        int temp = allNodes.get(low).val;
        allNodes.get(low).val = allNodes.get(high).val;
        allNodes.get(high).val = temp;
    }

    /**
     * 中序遍历
     *
     * @param root
     * @return
     */
    private static List<TreeNode> inOrder(TreeNode root) {
        List<TreeNode> nodes = new ArrayList<>();
        if (root != null) {
            nodes.addAll(inOrder(root.left));
            nodes.add(root);
            nodes.addAll(inOrder(root.right));
        }
        return nodes;
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(3);
        root.left = new TreeNode(1);
        root.right = new TreeNode(4);
        root.right.left = new TreeNode(2);

        recoverTree(root);
        System.out.println("恢复之前");
        root.print();
        System.out.println();
        System.out.println("恢复之后");
        root.print();
    }
}

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