标签：style blog http color 使用 io strong ar for
LeetCode: 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?
地址：https://oj.leetcode.com/problems/recover-binary-search-tree/
算法：二叉搜索树中有两个节点被错误的交换了，要求找到这两个节点，并把二者交换回来，要求用常数的空间。首先，使用中序遍历来搜索二叉树，当找到一个树比他的前趋还要小的时候，说明这个点是第一个错误点，即为A，然后接下去找第二个错误点，第二个错误点应该在第一个大于A点值的前面一个节点，记为B，接下去交换A，B节点即可。注意处理B点为最后一个节点时的情况。代码：
/**
* 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) {
if(!root) return ;
stack<TreeNode *> stk;
TreeNode *p = root;
while(p){
stk.push(p);
p = p->left;
}
TreeNode *pre = NULL;
p = NULL;
TreeNode *swapped_node1 = NULL;
bool is_swapped = false;
bool find_node = false;
while(!stk.empty()){
pre = p;
p = stk.top();
stk.pop();
if(!find_node && pre && pre->val > p->val){
swapped_node1 = pre;
find_node = true;
}
if(find_node && p->val >= swapped_node1->val){
if(pre){
int temp = swapped_node1->val;
swapped_node1->val = pre->val;
pre->val = temp;
is_swapped = true;
break;
}
}
if(p->right){
TreeNode *q = p->right;
while(q){
stk.push(q);
q = q->left;
}
}
}
if(!is_swapped){
int temp = swapped_node1->val;
swapped_node1->val = p->val;
p->val = temp;
}
return ;
}
};
LeetCode: Recover Binary Search Tree
标签：style blog http color 使用 io strong ar for
原文地址：http://www.cnblogs.com/boostable/p/leetcode_recover_binary_search_tree.html
