【LeetCode 热题 100】二叉树的最大深度 / 翻转二叉树 / 二叉树的直径 / 验证二叉搜索树

二叉树的中序遍历
- 二叉树的中序遍历
class Solution {vector<int> res;
public:vector<int> inorderTraversal(TreeNode* root) {dfs(root);return res;}void dfs(TreeNode* root){if (root == nullptr) return;dfs(root->left);res.push_back(root->val);dfs(root->right);}
};
二叉树的最大深度
- 二叉树的最大深度
class Solution {
public:int maxDepth(TreeNode* root) {if (root == nullptr) return 0;int left = maxDepth(root->left);int right = maxDepth(root->right);return left > right ? left + 1 : right + 1;}
};
翻转二叉树
- 翻转二叉树
class Solution {
public:TreeNode* invertTree(TreeNode* root) {if (root == nullptr) return nullptr;TreeNode *left = invertTree(root->left);TreeNode *right = invertTree(root->right);root->left = right;root->right = left;return root;}
};
对称二叉树
- 对称二叉树
class Solution {
public:bool isSymmetric(TreeNode* root) {return dfs(root->left, root->right);}bool dfs(TreeNode* left, TreeNode* right){if (left && right){if (left->val != right->val) return false;return dfs(left->left, right->right) && dfs(left->right, right->left);}else if (left != right) return false;else return true;}
};
二叉树的直径
- 二叉树的直径
class Solution {int depth;
public:int diameterOfBinaryTree(TreeNode* root) {dfs(root);return depth - 1;}int dfs(TreeNode* root){if (root == nullptr) return 0;int left = dfs(root->left);int right = dfs(root->right);depth = max(depth, left + right + 1);return max(left, right) + 1;}
};
二叉树的层序遍历
- 二叉树的层序遍历
class Solution {
public:vector<vector<int>> levelOrder(TreeNode* root) {vector<vector<int>> res;queue<TreeNode*> q;if (root == nullptr) return res;q.push(root);while (q.size()){int sz = q.size();vector<int> tmp;while (sz--){TreeNode *node = q.front();tmp.push_back(node->val);q.pop();if (node->left) q.push(node->left);if (node->right) q.push(node->right);}res.push_back(tmp);}return res;}
};
将有序数组转换为二叉搜索树
- 将有序数组转换为二叉搜索树
class Solution {
public:TreeNode* sortedArrayToBST(vector<int>& nums) {return dfs(nums, 0, nums.size() - 1);}TreeNode* dfs(vector<int>& nums, int l, int r){if (l > r) return nullptr;int mid = l + (r - l) / 2;TreeNode* node = new TreeNode(nums[mid]);node->left = dfs(nums, l, mid - 1);node->right = dfs(nums, mid + 1, r);return node;}
};
验证二叉搜索树
- 验证二叉搜索树
递归遍历。
class Solution {
public:bool isValidBST(TreeNode* root) {return dfs(root, LONG_MIN, LONG_MAX);}bool dfs(TreeNode* root, long min_val, long max_val){if (root == nullptr) return true;if (root->val <= min_val || root->val >= max_val) return false;return dfs(root->left, min_val, root->val) && dfs(root->right, root->val, max_val);}
};
前序遍历。
class Solution {long prev = LONG_MIN;
public:bool isValidBST(TreeNode* root) {if (root == nullptr) return true;if (isValidBST(root->left) == false) return false;if (root->val <= prev) return false;prev = root->val; return isValidBST(root->right);}
};
二叉搜索树中第 K 小的元素
- 二叉搜索树中第 K 小的元素
class Solution {int res, cnt;
public:int kthSmallest(TreeNode* root, int k) {cnt = k;dfs(root);return res;}void dfs(TreeNode* root){if (root == nullptr) return;dfs(root->left);if (--cnt == 0) {res = root->val;return;}dfs(root->right);}
};
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