代码随想录二刷——二叉树day15

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前言

一一个本硕双非的小菜鸡，备战24年秋招，计划二刷完卡子哥的刷题计划，加油！
二刷决定精刷了，于是参加了卡子哥的刷题班，训练营为期60天，我一定能坚持下去，迎来两个月后的脱变的，加油！
推荐一手卡子哥的刷题网站，感谢卡子哥。代码随想录

一、102. 二叉树的层序遍历

102. 二叉树的层序遍历
Note：层序遍历又称广度优先搜索，一层层的将数据打印出来，使用的是队列操作。

/**
 * 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) {
        queue<TreeNode*> que;

        if (root != NULL) que.push(root);
        vector<vector<int>> result;

        while (!que.empty()) {
            int size = que.size();
            vector<int> vec;

            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;
    }
};

Note：递归法

/**
 * 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:
    void order(TreeNode* cur, vector<vector<int>>& result, int depth) {
        if (cur == NULL) return;
        if (result.size() == depth) result.push_back(vector<int>());
        result[depth].push_back(cur->val);
        order(cur->left, result, depth + 1);
        order(cur->right, result, depth + 1);
    }
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> result;
        int depth = 0;
        order(root, result, depth);
        return result;
    }
};

二、107. 二叉树的层序遍历 II

107. 二叉树的层序遍历 II

Note：正常遍历之后翻转一下

/**
 * 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>> levelOrderBottom(TreeNode* root) {
        queue<TreeNode*> que;
        if (root != NULL) que.push(root);
        vector<vector<int>> result;

        while (!que.empty()) {
            int size = que.size();
            vector<int> vec;

            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);
        }
        reverse(result.begin(), result.end());
        return result;
    }
};

Note：递归法

/**
 * 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:
    void order(TreeNode* cur, vector<vector<int>>& result, int depth) {
        if (cur == NULL) return;
        if (result.size() == depth) result.push_back(vector<int>());
        result[depth].push_back(cur->val);
        order(cur->left, result, depth + 1);
        order(cur->right, result, depth + 1);
    }
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> result;
        int depth = 0;
        order(root, result, depth);
        reverse(result.begin(), result.end());
        return result;
    }
};

三、637. 二叉树的层平均值

637. 二叉树的层平均值

Note：就是层序遍历加了个求平均

/**
 * 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<double> averageOfLevels(TreeNode* root) {
        queue<TreeNode*> que;
        if (root != NULL) que.push(root);
        vector<double> result;

        while (!que.empty()) {
            int size = que.size();
            double sum = 0;

            for (int i = 0; i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                sum += node->val;
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            result.push_back(sum / size);
        }
        return result;
    }
};

四、429. N 叉树的层序遍历

429. N 叉树的层序遍历

Note：一样是模板题，只不过由left和right变成了children

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector children;

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        if (!root) {
            return {};
        }

        vector<vector<int>> ans;
        queue<Node*> q;
        q.push(root);

        while (!q.empty()) {
            int cnt = q.size();
            vector<int> level;
            for (int i = 0; i < cnt; ++i) {
                Node* cur = q.front();
                q.pop();
                level.push_back(cur->val);
                for (Node* child: cur->children) {
                    q.push(child);
                }
            }
            ans.push_back(move(level));
        }

        return ans;
    }
};

第二种写法

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector children;

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        queue<Node*> que;
        if (root != NULL) que.push(root);
        vector<vector<int>> result;

        while (!que.empty()) {
            int size = que.size();
            vector<int> vec;
            for (int i = 0; i < size; i++) {
                Node* node = que.front();
                que.pop();
                vec.push_back(node->val)
                for (int i = 0; i < node->children.size(); i++) {
                    if (node->children[i]) que.push(node->children[i]);
                }
            }
            result.push_back(vec);
        }
        return result;
    }
};

五、199. 二叉树的右视图

Note：也是层序遍历的一种

/**
 * 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) {
        queue<TreeNode*> que;
        if (root != NULL) que.push(root);
        vector<int> result;

        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;
    }
};

六、515. 在每个树行中找最大值

Note：模板题+1，每次取个最大值就好了

/**
 * 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) {
        queue<TreeNode*> que;

        if (root != NULL) que.push(root);
        vector<vector<int>> result;

        while (!que.empty()) {
            int size = que.size();
            vector<int> vec;

            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;
    }
};

七、116. 填充每个节点的下一个右侧节点指针

Note：定义俩指针，记录并传递

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/

class Solution {
public:
    Node* connect(Node* root) {
        queue<Node*> que;
        if (root != NULL) que.push(root);
        while (!que.empty()) {
            int size = que.size();
            Node* nodePre;
            Node* node;

            for (int i = 0; i < size; i++) {
                if (i == 0) {
                    nodePre = que.front();
                    que.pop();
                    node = nodePre;
                } else {
                    node = que.front();
                    que.pop();
                    nodePre->next = node;
                    nodePre = nodePre->next;
                }
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            nodePre->next = NULL;
        }
        return root;
    }
};

八、117. 填充每个节点的下一个右侧节点指针 II

Note：定义俩指针，记录并传递

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/

class Solution {
public:
    Node* connect(Node* root) {
        queue<Node*> que;
        if (root != NULL) que.push(root);
        while (!que.empty()) {
            int size = que.size();
            vector<int> vet;
            Node* nodePre;
            Node* node;

            for (int i = 0; i < size; i++) {
                if (i == 0) {
                    nodePre = que.front();
                    que.pop();
                    node = nodePre;
                } else {
                    node = que.front();
                    que.pop();
                    nodePre->next = node;
                    nodePre = nodePre->next;
                }
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            nodePre->next = NULL;
        }
        return root;
    }
};

九、104. 二叉树的最大深度

Note：也是一道模板题

/**
 * 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 == NULL) return 0;
        int depth = 0;

        queue<TreeNode*> que;
        que.push(root);

        while(!que.empty()) {
            int size = que.size();
            depth++;

            for (int i = 0; i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
        }
        return depth;
    }
};

十、111. 二叉树的最小深度

Note：只需要在104题的基础上增加左右孩子都为空的判断就成

/**
 * 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 minDepth(TreeNode* root) {
        if(root == NULL) return 0;
        int depth = 0;
        queue<TreeNode*> que;
        que.push(root);

        while (!que.empty()) {
            int size = que.size();
            depth++;

            for (int i = 0;i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
                if (!node->left && !node->right) return depth;
            }
        }
        return depth;
    }
};

十一、226. 翻转二叉树

226. 翻转二叉树
Note：递归法

/**
 * 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 == NULL) return root;
        swap(root->left, root->right);
        invertTree(root->left);
        invertTree(root->right);
        return root;
    }
};

十二、101. 对称二叉树

101. 对称二叉树
Note：递归法

/**
 * 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 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;
        else {
            bool outside = compare(left->left, right->right);
            bool inside = compare(left->right, right->left);
            bool isSame = outside && inside;
            return isSame;
        }
    }
    bool isSymmetric(TreeNode* root) {
        if (root == NULL) return true;
        return compare(root->left, root->right);
    }
};

总结

我要打十个！