99. 恢复二叉搜索树

English Version

题目描述

给你二叉搜索树的根节点 root ，该树中的两个节点被错误地交换。请在不改变其结构的情况下，恢复这棵树。

进阶：使用 O(n) 空间复杂度的解法很容易实现。你能想出一个只使用常数空间的解决方案吗？

 

示例 1：

输入：root = [1,3,null,null,2]
输出：[3,1,null,null,2]
解释：3 不能是 1 左孩子，因为 3 > 1 。交换 1 和 3 使二叉搜索树有效。

示例 2：

输入：root = [3,1,4,null,null,2]
输出：[2,1,4,null,null,3]
解释：2 不能在 3 的右子树中，因为 2 < 3 。交换 2 和 3 使二叉搜索树有效。

 

提示：

• 树上节点的数目在范围 [2, 1000] 内
• -231 <= Node.val <= 231 - 1

解法

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);
        }
    }
}

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