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?
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { private: void swap(int& a,int&b) { int tmp=a; a=b; b=tmp; } TreeNode* findlarge(TreeNode* root,int val) { if(root==NULL) return NULL; TreeNode* result=NULL; if(root->val>val) { val=root->val; result=root; } TreeNode* p1=findlarge(root->left,val); TreeNode* p2=findlarge(root->right,val); if(p1!=NULL) result=p1; if(p2!=NULL) { if(result==NULL) result=p2; else if(result->val<p2->val) result=p2; } return result; } TreeNode* findsmall(TreeNode* root,int val) { if(root==NULL) return NULL; TreeNode* result=NULL; if(root->val<val) { val=root->val; result=root; } TreeNode* p1=findsmall(root->left,val); TreeNode* p2=findsmall(root->right,val); if(p1!=NULL) result=p1; if(p2!=NULL) { if(result==NULL) result=p2; else if(result->val>p2->val) result=p2; } return result; } public: void recoverTree(TreeNode *root) { if(root==NULL) return; TreeNode* left=findlarge(root->left,root->val); TreeNode* right=findsmall(root->right,root->val); if(left!=NULL && right!=NULL) { swap(left->val,right->val); return; } if(right!=NULL) { swap(root->val,right->val); return; } if(left!=NULL) { swap(root->val,left->val); return; } recoverTree(root->left); recoverTree(root->right); } };
Recover Binary Search Tree,布布扣,bubuko.com
原文:http://www.cnblogs.com/erictanghu/p/3759585.html