首页 > 代码库 > leetcode 之 Recover Binary Search Tree

leetcode 之 Recover Binary Search Tree

2024-07-17 20:07:36   221人阅读

Recover Binary Search Tree

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

思路：二叉树的中序遍历，如果某个节点的前序节点大于该节点，则对于第一个出错位置来说，前序是错误位置；对于第二个出错位置来说，后续是出错位置，题目要求不要申请空间，所以要用递归，下面我们用递归和栈都实现一下：

class Solution {
public:
    void recoverTree(TreeNode *root) {
        if(root == NULL)return;
	TreeNode* pre = NULL,*n1 = NULL,*n2 = NULL;
	findTwoNode(root,n1,n2,pre);
	if(n1 && n2)
	{
		int temp = n1->val;
		n1->val = n2->val;
		n2->val = temp;
	}
    }
    void findTwoNode(TreeNode* root,TreeNode* &n1,TreeNode* &n2,TreeNode* &pre)
{
	if(root == NULL)return;
	findTwoNode(root->left,n1,n2,pre);
	if(pre && pre->val > root->val)
	{
		n2 = root;//第二个出错位置是该节点的后序
		if(n1 == NULL)
		{
			n1 = pre;//第一个出错位置是该节点的先序
		}
	}
	pre = root;
	findTwoNode(root->right,n1,n2,pre);
}
};
下面是堆栈实现，思路一样：

struct TreeNode {
     int val;
     TreeNode *left;
     TreeNode *right;
     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    void recoverTree(TreeNode *root) {
    	if(!root)return;
    	stack<TreeNode*> s;
    	TreeNode* p = root , *pre = NULL ,*node1 = NULL,*node2 = NULL;
    	while(p || !s.empty())
    	{
    		while( p )
    		{
    			s.push(p);
    			p = p -> left;
    		}
    		p = s.top();
    		s.pop();
    		if(pre && pre -> val > p -> val)
    		{
    			if( !node1 )node1 = pre;
    			node2 = p;
    		}
    		pre = p;
    		p = p -> right;
    	}
    	if(node1 && node2) swap(node1 -> val,node2 -> val);
    }
};


联系
我们