README_EN.md

99. Recover Binary Search Tree

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Description

You are given the root of a binary search tree (BST), where the values of exactly two nodes of the tree were swapped by mistake. Recover the tree without changing its structure.

 

Example 1:

Input: root = [1,3,null,null,2]
Output: [3,1,null,null,2]
Explanation: 3 cannot be a left child of 1 because 3 > 1. Swapping 1 and 3 makes the BST valid.

Example 2:

Input: root = [3,1,4,null,null,2]
Output: [2,1,4,null,null,3]
Explanation: 2 cannot be in the right subtree of 3 because 2 < 3. Swapping 2 and 3 makes the BST valid.

 

Constraints:

• The number of nodes in the tree is in the range [2, 1000].
• -231 <= Node.val <= 231 - 1

 

Follow up: A solution using O(n) space is pretty straight-forward. Could you devise a constant O(1) space solution?

Solutions

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def recoverTree(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """

        def dfs(root):
            nonlocal prev, first, second
            if root:
                dfs(root.left)
                if prev:
                    if first is None and root.val < prev.val:
                        first = prev
                    if first and root.val < prev.val:
                        second = root
                prev = root
                dfs(root.right)

        prev = first = second = None
        dfs(root)
        first.val, second.val = second.val, first.val

Java

/**
 * 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 {
    private TreeNode prev;
    private TreeNode first;
    private TreeNode second;

    public void recoverTree(TreeNode root) {
        dfs(root);
        int t = first.val;
        first.val = second.val;
        second.val = t;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        dfs(root.left);
        if (prev != null) {
            if (first == null && prev.val > root.val) {
                first = prev;
            }
            if (first != null && prev.val > root.val) {
                second = root;
            }
        }
        prev = root;
        dfs(root.right);
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* prev;
    TreeNode* first;
    TreeNode* second;

    void recoverTree(TreeNode* root) {
        dfs(root);
        swap(first->val, second->val);
    }

    void dfs(TreeNode* root) {
        if (!root) return;
        dfs(root->left);
        if (prev) {
            if (!first && prev->val > root->val) first = prev;
            if (first && prev->val > root->val) second = root;
        }
        prev = root;
        dfs(root->right);
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func recoverTree(root *TreeNode) {
	var prev *TreeNode
	var first *TreeNode
	var second *TreeNode

	var dfs func(root *TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		dfs(root.Left)
		if prev != nil {
			if first == nil && prev.Val > root.Val {
				first = prev
			}
			if first != nil && prev.Val > root.Val {
				second = root
			}
		}
		prev = root
		dfs(root.Right)
	}

	dfs(root)
	first.Val, second.Val = second.Val, first.Val
}

C#

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    private TreeNode prev, first, second;

    public void RecoverTree(TreeNode root) {
        dfs(root);
        int t = first.val;
        first.val = second.val;
        second.val = t;
    }

    private void dfs(TreeNode root) {
        if (root != null)
        {
            dfs(root.left);
            if (prev != null)
            {
                if (first == null && prev.val > root.val)
                {
                    first = prev;
                }
                if (first != null && prev.val > root.val)
                {
                    second = root;
                }
            }
            prev = root;
            dfs(root.right);
        }
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {void} Do not return anything, modify root in-place instead.
 */
const recoverTree = root => {
    const data = {
        prev: null,
        first: null,
        second: null,
    };
    let tmp = 0;

    helper(root, data);

    tmp = data.first.val;
    data.first.val = data.second.val;
    data.second.val = tmp;
};

const helper = (root, data) => {
    if (!root) return;

    helper(root.left, data);

    if (data.prev && data.prev.val >= root.val) {
        if (!data.first) data.first = data.prev;
        data.second = root;
    }

    data.prev = root;

    helper(root.right, data);
};

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