Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Note:
confused what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.
给定的二叉搜索树中有两个节点的值错换了，找出这两个节点，恢复二叉搜索树。要求不适用额外的空间。
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void recoverTree(TreeNode *root) {
stack<TreeNode*> st;
TreeNode* pointer=root;
TreeNode* prev=NULL;
TreeNode* nodeLarge=NULL;
TreeNode* nodeSmall=NULL;
while(pointer){st.push(pointer); pointer=pointer->left;}
while(!st.empty()){
TreeNode* cur = st.top();
st.pop();
if(prev && prev->val > cur->val){
if(nodeLarge==NULL || prev->val > nodeLarge->val) nodeLarge=prev;
if(nodeSmall==NULL || cur->val < nodeSmall->val) nodeSmall=cur;
}
prev=cur;
pointer=cur->right;
while(pointer){st.push(pointer); pointer=pointer->left;}
}
//替换两个节点的值
int temp=nodeLarge->val;
nodeLarge->val = nodeSmall->val;
nodeSmall->val = temp;
}
};LeetCode: Recover Binary Search Tree [099],布布扣,bubuko.com
LeetCode: Recover Binary Search Tree [099]
原文地址：http://blog.csdn.net/harryhuang1990/article/details/28092393
