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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.
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); } };
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