An example of in-order traversal application. My intuition is that, we have to serialize it into an array and check, but in-order traversal does exactly the same thing. Then, you only need bookmark some runtime records to get it accepted :)
class Solution { public: vector<TreeNode *> rec; queue<TreeNode *> q; void go(TreeNode *root) { if(root->left) go(root->left); // update record if(q.size() == 2) q.pop(); q.push(root); // check if(q.size() > 1) { int v1 = q.back()->val; int v0 = q.front()->val; if(v0 > v1) { if(rec.empty()) { rec.push_back(q.front()); rec.push_back(q.back()); } else { rec[1] = q.back(); } } } if(root->right) go(root->right); } void recoverTree(TreeNode *root) { if (!root) return; if(!root->left && !root->right) return; go(root); int tmp = rec[0]->val; rec[0]->val = rec[1]->val; rec[1]->val = tmp; } };
LeetCode "Recover Binary Search Tree",布布扣,bubuko.com
LeetCode "Recover Binary Search Tree"
原文：http://www.cnblogs.com/tonix/p/3891384.html