Question

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 恰好 两个节点的值被错误地交换。_请在不改变其结构的情况下，恢复这棵树 _。

 

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 的左孩子，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

 

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 O(1) 空间的解决方案吗？

解法

方法一：中序遍历

中序遍历二叉搜索树，得到的序列是递增的。如果有两个节点的值被错误地交换，那么中序遍历得到的序列中，一定会出现两个逆序对。我们用 first 和 second 分别记录这两个逆序对中较小值和较大值的节点，最后交换这两个节点的值即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉搜索树的节点个数。

# 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

/**
 * 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);
  }
}

/**
 * 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);
  }
};

/**
 * 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
}

/**
 * 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;
};

/**
 * 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);
  }
}