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LeetCode Hot 100：二叉树

article 2024/10/28 7:37:14

LeetCode Hot 100：二叉树

94. 二叉树的中序遍历

思路 1：递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
private:
    vector<int> ans;

public:
    vector<int> inorderTraversal(TreeNode* root) {
        helper(root);
        return ans;
    }
    void helper(TreeNode* root) {
        if (root == nullptr)
            return;

        helper(root->left);
        ans.push_back(root->val);
        helper(root->right);
    }
};

104. 二叉树的最大深度

思路 1：递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode* root) {
        return root ? max(maxDepth(root->left), maxDepth(root->right)) + 1 : 0;
    }
};

思路 2：层序遍历

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode* root) {
        if (root == nullptr)
            return 0;

        queue<TreeNode*> q;
        q.push(root);
        int depth = 0;
        while (!q.empty()) {
            int sz = q.size();
            for (int i = 0; i < sz; i++) {
                TreeNode* node = q.front();
                q.pop();
                if (node->left)
                    q.push(node->left);
                if (node->right)
                    q.push(node->right);
            }
            depth++;
        }
        return depth;
    }
};

226. 翻转二叉树

思路 1：自底向上递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if (root == nullptr)
            return root;

        swap(root->left, root->right);
        invertTree(root->left);
        invertTree(root->right);
        return root;
    }
};

思路 2：自顶向下递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if (root == nullptr)
            return root;

        TreeNode* l = invertTree(root->left);
        TreeNode* r = invertTree(root->right);
        root->left = r;
        root->right = l;

        return root;
    }
};

101. 对称二叉树

思路 1：递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        return isSymmetric(root->left, root->right);
    }
    bool isSymmetric(TreeNode* left, TreeNode* right) {
        if (left == nullptr && right == nullptr)
            return true;
        if ((left && right == nullptr) || (left == nullptr && right))
            return false;

        if (left->val != right->val)
            return false;

        return isSymmetric(left->left, right->right) &&
               isSymmetric(left->right, right->left);
    }
};

思路 2：迭代

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) { return isSymmetric(root, root); }
    bool isSymmetric(TreeNode* u, TreeNode* v) {
        queue<TreeNode*> q;
        q.push(u);
        q.push(v);
        while (!q.empty()) {
            u = q.front();
            q.pop();
            v = q.front();
            q.pop();
            if (u == nullptr && v == nullptr)
                continue;
            if ((u && v == nullptr) || (u == nullptr && v))
                return false;
            if (u->val != v->val)
                return false;

            q.push(u->left);
            q.push(v->right);

            q.push(u->right);
            q.push(v->left);
        }

        return true;
    }
};

543. 二叉树的直径

思路 1：递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    int diameterOfBinaryTree(TreeNode* root) {
        int ans = 0;

        function<int(TreeNode*)> dfs = [&](TreeNode* root) {
            if (root == nullptr)
                return 0;

            int left = dfs(root->left);
            int right = dfs(root->right);
            ans = max(ans, left + right + 1);
            return max(left, right) + 1;
        };

        dfs(root);

        return ans - 1;
    }
};

102. 二叉树的层序遍历

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        if (root == nullptr)
            return {};

        queue<TreeNode*> q;
        q.push(root);
        vector<vector<int>> ans;
        while (!q.empty()) {
            int sz = q.size();
            vector<int> v;
            for (int i = 0; i < sz; i++) {
                TreeNode* node = q.front();
                q.pop();
                v.push_back(node->val);
                if (node->left)
                    q.push(node->left);
                if (node->right)
                    q.push(node->right);
            }
            ans.push_back(v);
        }
        return ans;
    }
};

108. 将有序数组转换为二叉搜索树

思路 1：递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return buildBST(nums, 0, nums.size() - 1);
    }
    TreeNode* buildBST(vector<int>& nums, int start, int end) {
        if (start > end)
            return nullptr;

        int mid = start + (end - start) / 2;
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = buildBST(nums, start, mid - 1);
        root->right = buildBST(nums, mid + 1, end);

        return root;
    }
};

98. 验证二叉搜索树

思路 1：递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        return isValidBST(root, LONG_LONG_MIN, LONG_LONG_MAX);
    }
    bool isValidBST(TreeNode* root, long long lower, long long upper) {
        if (root == nullptr)
            return true;
        if (root->val <= lower)
            return false;
        if (root->val >= upper)
            return false;
        return isValidBST(root->left, lower, root->val) &&
               isValidBST(root->right, root->val, upper);
    }
};

思路 2：中序遍历

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
private:
    vector<int> arr;

public:
    bool isValidBST(TreeNode* root) {
        inOrder(root);
        for (int i = 0; i < arr.size() - 1; i++)
            if (arr[i] >= arr[i + 1])
                return false;
        return true;
    }
    void inOrder(TreeNode* root) {
        if (root == nullptr)
            return;

        if (root->left)
            inOrder(root->left);
        arr.push_back(root->val);
        if (root->right)
            inOrder(root->right);
    }
};

230. 二叉搜索树中第 K 小的元素

思路 1：中序遍历 + 辅助数组

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
private:
    vector<int> arr;

public:
    int kthSmallest(TreeNode* root, int k) {
        inOrder(root);
        
        if (k > arr.size())
            return -1;
        
        return arr[k - 1];
    }
    void inOrder(TreeNode* root) {
        if (root == nullptr)
            return;

        if (root->left)
            inOrder(root->left);
        arr.push_back(root->val);
        if (root->right)
            inOrder(root->right);
    }
};

思路 2：中序遍历

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
private:
    int ans = 0;

public:
    int kthSmallest(TreeNode* root, int k) {
        inOrder(root, k);
        return ans;
    }
    // 辅函数 - 中序遍历
    void inOrder(TreeNode* root, int& k) {
        if (root == nullptr)
            return;
        inOrder(root->left, k);
        k--;
        if (k == 0) {
            ans = root->val;
            return;
        }
        if (k > 0)
            inOrder(root->right, k);
    }
};

199. 二叉树的右视图

思路 1：层序遍历

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        if (root == nullptr)
            return {};

        queue<TreeNode*> q;
        q.push(root);
        vector<int> ans;
        while (!q.empty()) {
            int sz = q.size();
            vector<int> v;
            for (int i = 0; i < sz; i++) {
                TreeNode* node = q.front();
                q.pop();
                v.push_back(node->val);
                if (node->left)
                    q.push(node->left);
                if (node->right)
                    q.push(node->right);
            }
            ans.push_back(v.back());
        }
        return ans;
    }
};

114. 二叉树展开为链表

思路 1：先序遍历 + 辅助数组

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
private:
    vector<int> arr;

public:
    void flatten(TreeNode* root) {
        if (root == nullptr)
            return;
        if (root->left == nullptr && root->right == nullptr)
            return;

        preOrder(root);
        root->left = nullptr;
        root->right = nullptr;
        TreeNode* p = root;
        for (int i = 1; i < arr.size(); i++) {
            TreeNode* node = new TreeNode(arr[i]);
            p->right = node;
            p = p->right;
        }
    }
    void preOrder(TreeNode* root) {
        if (root == nullptr)
            return;

        arr.push_back(root->val);
        if (root->left)
            preOrder(root->left);
        if (root->right)
            preOrder(root->right);
    }
};

105. 从前序与中序遍历序列构造二叉树

思路 1：递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution
{
private:
    unordered_map<int, int> index;

public:
    // 主函数
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder)
    {
        int n = preorder.size();
        // 构造哈希映射，帮助我们快速定位根节点
        for (int i = 0; i < n; i++)
            index[inorder[i]] = i;
        return buildTree(preorder, inorder, 0, n - 1, 0, n - 1);
    }
    // 辅函数
    TreeNode *buildTree(const vector<int> &preorder, const vector<int> &inorder, int preorder_left, int preorder_right, int inorder_left, int inorder_right)
    {
        if (preorder_left > preorder_right)
            return nullptr;
        // 前序遍历中的第一个节点就是根节点
        int preorder_root = preorder_left;
        // 在中序遍历中定位根节点
        int inorder_root = index[preorder[preorder_root]];
        // 先把根节点建立出来
        TreeNode *root = new TreeNode(preorder[preorder_root]);
        // 得到左子树中的节点数目
        int size_left_subtree = inorder_root - inorder_left;
        // 递归地构造左子树，并连接到根节点
        // 先序遍历中「从 左边界+1 开始的 size_left_subtree」个元素就对应了中序遍历中「从 左边界 开始到 根节点定位-1」的元素
        root->left = buildTree(preorder, inorder, preorder_left + 1, preorder_left + size_left_subtree, inorder_left, inorder_root - 1);
        // 递归地构造右子树，并连接到根节点
        // 先序遍历中「从 左边界+1+左子树节点数目 开始到 右边界」的元素就对应了中序遍历中「从 根节点定位+1 到 右边界」的元素
        root->right = buildTree(preorder, inorder, preorder_left + size_left_subtree + 1, preorder_right, inorder_root + 1, inorder_right);
        return root;
    }
};

437. 路径总和 III

思路 1：递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    int pathSum(TreeNode* root, int targetSum) {
        if (root == nullptr)
            return 0;

        int res = path_sum(root, targetSum);
        res += pathSum(root->left, targetSum);
        res += pathSum(root->right, targetSum);

        return res;
    }
    long long path_sum(TreeNode* root, long long target) {
        if (root == nullptr)
            return 0LL;

        long long count = root->val == target ? 1LL : 0LL;
        target -= root->val;
        count += path_sum(root->left, target);
        count += path_sum(root->right, target);

        return count;
    }
};

思路 2：回溯 + 哈希表

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
private:
    unordered_map<long long, int> prefix; // <总和，次数>

public:
    int pathSum(TreeNode* root, int targetSum) {
        prefix[0] = 1; // 初始化
        return backtrack(root, 0LL, targetSum);
    }
    // 辅函数
    int backtrack(TreeNode* root, long long cur, int target) {
        if (root == nullptr)
            return 0;

        int res = 0;
        cur += root->val;
        if (prefix.count(cur - target))
            res += prefix[cur - target];
        prefix[cur]++; // 修改当前节点状态
        // 递归子节点
        res += backtrack(root->left, cur, target);
        res += backtrack(root->right, cur, target);
        prefix[cur]--; // 回改当前节点状态
        
        return res;
    }
};

236. 二叉树的最近公共祖先

思路 1：递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (root == nullptr)
            return nullptr;
        // p 和 q 其中有一个正好是 root，直接返回 root
        if (root == p || root == q)
            return root;
        // 通过递归，得到左右两棵子树的值
        TreeNode* leftLCA = lowestCommonAncestor(root->left, p, q);
        TreeNode* rightLCA = lowestCommonAncestor(root->right, p, q);
        // p 和 q 分别在 root 的不同子树，直接返回 root
        if (leftLCA && rightLCA)
            return root;
        // p 和 q 在 root 的同一侧，且 root 不等于 p 或者 q 的任何一个，那么就找 p 和 q 在的那一侧子树
        return leftLCA == nullptr ? rightLCA : leftLCA;
    }
};

124. 二叉树中的最大路径和

思路 1：递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    int maxPathSum(TreeNode* root) {
        int ans = INT_MIN;

        function<int(TreeNode*)> dfs = [&](TreeNode* root) -> int {
            if (root == nullptr)
                return 0;

            int left = dfs(root->left);
            int right = dfs(root->right);
            int sum = root->val + left + right;
            ans = max(ans, sum);

            return max(max(left, right) + root->val, 0);
        };

        dfs(root);

        return ans;
    }
};

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