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):
            if root is None:
                return
            nonlocal prev, first, second
            dfs(root.left)
            if prev and prev.val > root.val:
                if first is None:
                    first = prev
                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 && prev.val > root.val) {
            if (first == null) {
                first = prev;
            }
            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:
    void recoverTree(TreeNode* root) {
        TreeNode* prev = nullptr;
        TreeNode* first = nullptr;
        TreeNode* second = nullptr;
        function<void(TreeNode* root)> dfs = [&](TreeNode* root) {
            if (!root) return;
            dfs(root->left);
            if (prev && prev->val > root->val) {
                if (!first) first = prev;
                second = root;
            }
            prev = root;
            dfs(root->right);
        };
        dfs(root);
        swap(first->val, second->val);
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func recoverTree(root *TreeNode) {
	var prev, first, second *TreeNode
	var dfs func(*TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		dfs(root.Left)
		if prev != nil && prev.Val > root.Val {
			if first == nil {
				first = prev
			}
			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) {
            return;
        }
        dfs(root.left);
        if (prev != null && prev.val > root.val) {
            if (first == null) {
                first = prev;
            }
            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.
 */
var recoverTree = function (root) {
    let prev = null;
    let first = null;
    let second = null;
    function dfs(root) {
        if (!root) {
            return;
        }
        dfs(root.left);
        if (prev && prev.val > root.val) {
            if (!first) {
                first = prev;
            }
            second = root;
        }
        prev = root;
        dfs(root.right);
    }
    dfs(root);
    const t = first.val;
    first.val = second.val;
    second.val = t;
};

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