方法一:利用中序遍历可以将二叉搜索树转变为一个有序数组,遍历这个数组可以将最小绝对差找到。
方法二:在中序遍历过程中直接利用双指针算出最小绝对差。
class Solution {
public:
int res = INT32_MAX;
TreeNode* pre = nullptr;
void traversal(TreeNode* cur) {
if (cur == nullptr) return;
traversal(cur->left);
if (pre) res = min(res, cur->val - pre->val);
pre = cur;
traversal(cur->right);
return;
}
int getMinimumDifference(TreeNode* root) {
traversal(root);
return res;
}
};
class Solution {
private:
vector res;
int count = 0;
int maxCount = 0;
TreeNode* pre = nullptr;
void traversal(TreeNode* cur) {
if (cur == nullptr) return;
traversal(cur->left);
if (pre == nullptr) {
count = 1;
}else if (cur->val == pre->val) {
count++;
}else {
count = 1;
}
pre = cur;
if (count == maxCount) res.push_back(cur->val);
if (count > maxCount) {
maxCount = count;
res.clear();
res.push_back(cur->val);
}
traversal(cur->right);
return;
}
public:
vector findMode(TreeNode* root) {
traversal(root);
return res;
}
};
从下到上处理一定要用后序遍历。
class Solution {
private:
TreeNode* traversal(TreeNode* root, TreeNode* p, TreeNode* q) {
if (root == NULL) return NULL;
if (root == p || root == q) return root;
TreeNode* left = traversal(root->left, p, q);
TreeNode* right = traversal(root->right, p, q);
if (left && right) return root;
else if (left && !right) return left;
else if (!left && right) return right;
else return NULL;
}
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
return traversal(root, p, q);
}
};