【数据结构刷题专题】—— 二叉树

二叉树

二叉树刷题框架

二叉树的定义：

struct TreeNode {
	int val;
	TreeNode* left;
	TreeNode* right;
	TreeNode(int x) : val(x), left(NULL), right(NULL);
};

1 二叉树的遍历方式

【1】前序遍历

class Solution {
public:
    void traversal(TreeNode* node, vector<int>& vec) {
        if (node == NULL) return;
        vec.push_back(node->val);
        traversal(node->left, vec);
        traversal(node->right, vec);
    }
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> result;
        traversal(root, result);
        return result;
    }
};

【2】后序遍历

class Solution {
public:
    void traversal(TreeNode* node, vector<int>& vec) {
        if (node == NULL) return;
        traversal(node->left, vec);
        traversal(node->right, vec);
        vec.push_back(node->val);
    }
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> result;
        traversal(root, result);
        return result;
    }
};

【3】中序遍历

class Solution {
public:
    void traversal(TreeNode* node, vector<int>& vec) {
        if (node == NULL) return;
        traversal(node->left, vec);
        vec.push_back(node->val);
        traversal(node->right, vec);
    }
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> result;
        traversal(root, result);
        return result;
    }
};

【4】层序遍历

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> result;
        queue<TreeNode*> que;
        if (root != NULL) que.push(root);
        while (!que.empty()) {
            vector<int> vec;
            int size = que.size();
            for (int i = 0; i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                vec.push_back(node->val);
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            result.push_back(vec);
        }
        return result;
    }
};

2 二叉树的属性

【1】101. 对称二叉树

class Solution {
public:
    bool compare(TreeNode* left, TreeNode* right) {
        if (left != NULL && right == NULL) return false;
        else if (left == NULL && right != NULL) return false;
        else if (left == NULL && right == NULL) return true;
        else if (left->val != right->val) return false;
        bool outside = compare(left->left, right->right);
        bool inside = compare(left->right, right->left);
        return outside && inside;
    }
    bool isSymmetric(TreeNode* root) {
        if (root == NULL) return true;
        return compare(root->left, root->right);
    }
};

【2】104. 二叉树的最大深度
迭代法:

class Solution {
public:
    int maxDepth(TreeNode* root) {
        if (root == NULL) return 0;
        int depth = 0;
        queue<TreeNode*> que;
        que.push(root);
        while (!que.empty()) {
            int size = que.size();
            depth++;
            while (size--) {
                TreeNode* node = que.front();
                que.pop();
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
        }
        return depth;
    }
};

递归法：

class Solution {
public:
    int getDepth(TreeNode* node) {
        if (node == NULL) return 0;
        int left = getDepth(node->left);
        int right = getDepth(node->right);
        return 1 + max(left, right);
    }
    int maxDepth(TreeNode* root) {
        return getDepth(root);
    }
};

【3】111.二叉树的最小深度
递归法：

class Solution {
public:
    int getDepth(TreeNode* node) {
        if (node == NULL) return 0;
        int left = getDepth(node->left);
        int right = getDepth(node->right);
        if (node->left != NULL && node->right == NULL) return 1 + left;
        if (node->left == NULL && node->right != NULL) return 1 + right;
        return 1 + min(left, right);
    }
    int minDepth(TreeNode* root) {
        return getDepth(root);
    }
};

迭代法：

class Solution {
public:
    int minDepth(TreeNode* root) {
        queue<TreeNode*> que;
        if (root == NULL) return 0;
        que.push(root);
        int depth = 0;
        while (!que.empty()) {
            int size = que.size();
            depth++;
            while (size--) {
                TreeNode* node = que.front();
                que.pop();
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
                if (node->left == NULL && node->right == NULL) return depth;
            }
        }
        return depth;
    }
};

【4】222. 完全二叉树的节点个数
递归法：

class Solution {
public:
    int getNum(TreeNode* node) {
        if (node == NULL) return 0;
        int left = getNum(node->left);
        int right = getNum(node->right);
        return 1 + left + right;
    }
    int countNodes(TreeNode* root) {
        return getNum(root);
    }
};

迭代法：

class Solution {
public:
    int countNodes(TreeNode* root) {
        queue<TreeNode*> que;
        if (root == NULL) return 0;
        que.push(root);
        int num = 0;
        while (!que.empty()) {
            int size = que.size();
            while (size--) {
                TreeNode* node = que.front();
                que.pop();
                num++;
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
        }
        return num;
    }
};

【5】110. 平衡二叉树

class Solution {
public:
    int getHeight(TreeNode* node) {
        if (node == NULL) return 0;
        int left = getHeight(node->left);
        if (left == -1) return -1;
        int right = getHeight(node->right);
        if (right == -1) return -1;
        return abs(left - right) > 1 ? -1 : 1 + max(left, right);
    }
    bool isBalanced(TreeNode* root) {
        return getHeight(root) == -1 ? false : true;
    }
};

【6】257. 二叉树的所有路径

class Solution {
public:
    void traversal(TreeNode* node, string path, vector<string>& result) {
        path += to_string(node->val);
        if (node->left == NULL && node->right == NULL) {
            result.push_back(path);
            return;
        }
        if (node->left) traversal(node->left, path + "->", result);
        if (node->right) traversal(node->right, path + "->", result);
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> result;
        string path;
        if (root == NULL) return result;
        traversal(root, path, result);
        return result;
    }
};

【7】404. 左叶子之和

class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        if (root == NULL) return 0;
        int left = sumOfLeftLeaves(root->left);
        if (root->left && root->left->left == NULL && root->left->right == NULL) {
            left = root->left->val;
        }
        int right = sumOfLeftLeaves(root->right);
        return left + right;
    }
};

【8】513. 找树左下角的值
迭代法：

class Solution {
public:
    int findBottomLeftValue(TreeNode* root) {
        queue<TreeNode*> que;
        que.push(root);
        int val = 0;
        while (!que.empty()) {
            int size = que.size();
            for (int i = 0; i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                if (i == 0) val = node->val;
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
        }
        return val;
    }
};

【9】112. 路径总和

class Solution {
public:
    bool pathSum(TreeNode* node, int count) {
        if (node->left == NULL && node->right == NULL && count == 0) return true;
        if (node->left == NULL && node->right == NULL) return false;
        if (node->left) {
            count -= node->left->val;
            if (pathSum(node->left, count)) return true;
            count += node->left->val;
        }
        if (node->right) {
            count -= node->right->val;
            if (pathSum(node->right, count)) return true;
            count += node->right->val;
        }
        return false;
    }
    bool hasPathSum(TreeNode* root, int targetSum) {
        if (root == NULL) return false;
        return pathSum(root, targetSum - root->val);
    }
};

【10】543. 二叉树的直径

class Solution {
public:
    int ans;
    int Depth(TreeNode* node) {
        if (node == NULL) return 0;
        int left = Depth(node->left);
        int right = Depth(node->right);
        ans = max(ans, 1 + left + right);
        return 1 + max(left, right);
    }
    int diameterOfBinaryTree(TreeNode* root) {
        ans = 1;
        Depth(root);
        return ans - 1;
    }
};

【11】124. 二叉树中的最大路径和

class Solution {
public:
    int ans = INT_MIN;
    int dfs(TreeNode* node) {
        if (node == NULL) return 0;
        ans = max(ans, node->val);
        int lSum = dfs(node->left);
        int rSum = dfs(node->right);
        lSum = max(0, lSum); rSum = max(0, rSum);
        ans = max(ans, node->val + lSum + rSum);
        return max(node->val + lSum, node->val + rSum);
    }
    int maxPathSum(TreeNode* root) {
        ans = max(ans, dfs(root));
        return ans;
    }
};

3 二叉树的修改和构造

【1】226. 翻转二叉树

class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if (root == NULL) return root;
        swap(root->left, root->right);
        if (root->left) invertTree(root->left);
        if (root->right) invertTree(root->right);
        return root;
    }
};

【2】106. 从中序与后序遍历序列构造二叉树

class Solution {
public:
    TreeNode* traversal(vector<int>& inorder, vector<int>& postorder) {
        if (postorder.size() == 0) return NULL;
        int rootValue = postorder[postorder.size() - 1];
        TreeNode* root = new TreeNode(rootValue);

        if (postorder.size() == 1) return root;
        int qiege;
        for (qiege = 0; qiege <= inorder.size(); qiege++) {
            if (inorder[qiege] == root->val) break;
        }
        vector<int> leftInorder(inorder.begin(), inorder.begin() + qiege);
        vector<int> rightInorder(inorder.begin() + qiege + 1, inorder.end());
        postorder.resize(postorder.size() - 1);
        vector<int> leftPostorder(postorder.begin(), postorder.begin() + leftInorder.size());
        vector<int> rightPostorder(postorder.begin() + leftInorder.size(), postorder.end());
        root->left = traversal(leftInorder, leftPostorder);
        root->right = traversal(rightInorder, rightPostorder);
        return root;
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        if (inorder.size() == 0 || postorder.size() == 0) return NULL;
        return traversal(inorder, postorder);
    }
};

【3】654. 最大二叉树
构造树一般采用的是前序遍历

class Solution {
public:
    TreeNode* traversal(vector<int>& nums, int left, int right) {
        if (left >= right) return NULL;
        int index = left;
        for (int i = left + 1; i < right; i++) {
            if (nums[i] > nums[index]) index = i;
        }
        TreeNode* root = new TreeNode(nums[index]);
        root->left = traversal(nums, left, index);
        root->right = traversal(nums, index + 1, right);
        return root;
    }
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        return traversal(nums, 0, nums.size());
    }
};

【4】617. 合并二叉树

class Solution {
public:
    TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
        if (root1 == NULL) return root2;
        if (root2 == NULL) return root1;
        root1->val += root2->val;
        root1->left = mergeTrees(root1->left, root2->left);
        root1->right = mergeTrees(root1->right, root2->right);
        return root1;
    }
};

【5】114. 二叉树展开为链表

class Solution {
public:
    void traversal(TreeNode* node) {
        if (node == NULL) return;
        traversal(node->left);
        traversal(node->right);
        TreeNode* left = node->left;
        TreeNode* right = node->right;
        node->left = NULL;
        node->right = left;
        while (node->right) node = node->right;
        node->right = right;
        return;
    }
    void flatten(TreeNode* root) {
        traversal(root);
        return;
    }
};

4 求二叉树的属性

【1】700. 二叉搜索树中的搜索

class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
        while (root != NULL) {
            if (root->val > val) {
                root = root->left;
            } else if (root->val < val) {
                root = root->right;
            } else return root;
        }
        return root;
    }
};

【2】98. 验证二叉搜索树

class Solution {
public:
    long long maxVel = LONG_MIN;
    bool isValidBST(TreeNode* root) {
        if (root == NULL) return true;
        bool left = isValidBST(root->left);
        if (root->val > maxVel) maxVel = root->val;
        else return false;
        bool right = isValidBST(root->right);
        return left && right;
    }
};

【3】530. 二叉搜索树的最小绝对差

把二叉搜索树转换成有序数组，然后遍历一遍数组，就统计出来最小差值

class Solution {
public:
    vector<int> vec;
    int ans = INT_MAX;
    void traversal(TreeNode* node) {
        if (node == NULL) return;
        traversal(node->left);
        vec.push_back(node->val);
        traversal(node->right);
    }
    int getMinimumDifference(TreeNode* root) {
        traversal(root);
        for (int i = 1; i < vec.size(); i++) {
            ans = min(ans, vec[i] - vec[i - 1]);
        }
        return ans;
    }
};

在递归中记录前一个节点的指针

class Solution {
public:
    int result = INT_MAX;
    TreeNode* pre = NULL;
    void traversal(TreeNode* node) {
        if (node == NULL) return;
        traversal(node->left);
        if (pre != NULL) result = min(result, node->val - pre->val);
        pre = node;
        traversal(node->right);
    }
    int getMinimumDifference(TreeNode* root) {
        traversal(root);
        return result;
    }
};

【4】501. 二叉搜索树中的众数

class Solution {
public:
    void traversal(TreeNode* cur, unordered_map<int, int>& map) {
        if (cur == NULL) return;
        map[cur->val]++;
        traversal(cur->left, map);
        traversal(cur->right, map);
        return;
    }
    static bool cmp(const pair<int, int>& a, const pair<int, int>& b) {
        return a.second > b.second;
    }
    vector<int> findMode(TreeNode* root) {
        unordered_map<int, int> map;
        vector<int> result;
        if (root == NULL) return result;
        traversal(root, map);
        vector<pair<int, int>> vec(map.begin(), map.end());
        sort(vec.begin(), vec.end(), cmp);
        result.push_back(vec[0].first);
        for (int i = 1; i < vec.size(); i++) {
            if (vec[0].second == vec[i].second) result.push_back(vec[i].first);
            else break;
        }
        return result;
    }
};

【5】把二叉搜索树转换为累加树

class Solution {
public:
    int pre = 0;
    void traversal(TreeNode* node) {
        if (node == NULL) return;
        traversal(node->right);
        node->val += pre;
        pre = node->val;
        traversal(node->left);

    }
    TreeNode* convertBST(TreeNode* root) {
        if (root == NULL) return root;
        traversal(root);
        return root;
    }
};

【6】230. 二叉搜索树中第K小的元素

class Solution {
public:
    int maxVel = INT_MIN;
    vector<int> vec;
    void traversal(TreeNode* node) {
        if (node == NULL) return;
        traversal(node->left);
        if (node->val > maxVel) {
            vec.push_back(node->val);
            maxVel = node->val;
        }
        traversal(node->right);
        return;
    }
    int kthSmallest(TreeNode* root, int k) {
        traversal(root);
        return vec[k - 1];
    }
};

【7】二叉树的右视图

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> result;
        queue<TreeNode*> que;
        if (root == NULL) return result;
        que.push(root);
        while (!que.empty()) {
            int size = que.size();
            for (int i = 0; i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                if (i == (size - 1)) result.push_back(node->val);
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
        }
        return result;
    }
};

5 二叉树的公共祖先问题

【1】236. 二叉树的最近公共祖先

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (root == p || root == q || root == NULL) return root;
        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);
        if (left != NULL && right != NULL) return root;
        if (left != NULL && right == NULL) return left;
        else if (left == NULL && right != NULL) return right;
        else return NULL;
    }
};

【2】235. 二叉搜索树的最近公共祖先

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        while (root) {
            if (root->val > q->val && root->val > p->val) root = root->left;
            else if (root->val < q->val && root->val < p->val) root = root->right;
            else return root;
        }
        return NULL;
    }
};

6 二叉搜索树的修改和构造

【1】二叉搜索树的插入操作

class Solution {
public:
    TreeNode* insertIntoBST(TreeNode* root, int val) {
        if (root == NULL) {
            TreeNode* node = new TreeNode(val);
            return node;
        }
        if (root->val > val) root->left = insertIntoBST(root->left, val);
        if (root->val < val) root->right = insertIntoBST(root->right, val);
        return root;
    }
};

【2】450. 删除二叉搜索树中的节点

class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (root == NULL) return root;
        if (root->val == key) {
            if (root->left == NULL && root->right == NULL) {
                delete root;
                return NULL;
            } else if (root->left == NULL) {
                auto tmp = root->right;
                delete root;
                return tmp;
            } else if (root->right == NULL) {
                auto tmp = root->left;
                delete root;
                return tmp;
            } else {
                TreeNode* cur = root->right;
                while (cur->left != NULL) {
                    cur = cur->left;
                }
                cur->left = root->left;
                TreeNode* tmp = root;
                root = root->right;
                delete tmp;
                return root;
            }
        }
        if (root->val > key) root->left = deleteNode(root->left, key);
        if (root->val < key) root->right = deleteNode(root->right, key);
        return root;
    }
};

【3】669. 修剪二叉搜索树

class Solution {
public:
    TreeNode* trimBST(TreeNode* root, int low, int high) {
        if (root == NULL) return root;
        if (root->val < low) return trimBST(root->right, low, high);
        if (root->val > high) return trimBST(root->left, low, high);
        root->left = trimBST(root->left, low, high);
        root->right = trimBST(root->right, low, high);
        return root;
    }
};

【4】108. 将有序数组转换为二叉搜索树

class Solution {
public:
    TreeNode* traversal(vector<int>& nums, int left, int right) {
        if (left > right) return NULL;
        int mid = left + (right - left) / 2;
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = traversal(nums, left, mid - 1);
        root->right = traversal(nums, mid + 1, right);
        return root;
    }
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return traversal(nums, 0, nums.size() - 1);
    }
};

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二叉树刷题专题到此结束，读者对题目有更好的解答欢迎讨论。