leetcode 之 Recover Binary Search Tree

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：二叉樹的中序遍歷，如果某個節點的前序節點大於該節點，則對於第一個出錯位置來說，前序是錯誤位置；對於第二個出錯位置來說，後續是出錯位置，題目要求不要申請空間，所以要用遞迴，下面我們用遞迴和棧都實現一下：

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;//第二個出錯位置是該節點的後序
		if(n1 == NULL)
		{
			n1 = pre;//第一個出錯位置是該節點的先序
		}
	}
	pre = root;
	findTwoNode(root->right,n1,n2,pre);
}
};

下面是堆疊實現，思路一樣：

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);
    }
};