【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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