代码随想录一刷打卡——二叉树（上篇）

---

前言

一个本硕双非的小菜鸡，备战24年秋招，计划刷完卡子哥的刷题计划，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、144. 二叉树的前序遍历

144. 二叉树的前序遍历
Note：中左右

/**
 * 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 travelsal (TreeNode* cur, vector<int>& vec) {
        if (cur == NULL) return;
        vec.push_back(cur->val);
        travelsal(cur->left, vec);
        travelsal(cur->right, vec);
    }
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> result;
        travelsal(root, result);
        return result;
    }
};

二、145. 二叉树的后序遍历

145. 二叉树的后序遍历

Note：左右中

/**
 * 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 tranversal(TreeNode* cur, vector<int>& vec) {
        if (cur == NULL) return;
        tranversal(cur->left, vec);
        tranversal(cur->right, vec);
        vec.push_back(cur->val);
    }
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> result;
        tranversal(root, result);
        return result;
    }
};

三、94. 二叉树的中序遍历

94. 二叉树的中序遍历

Note：左中右

/**
 * 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 traversal(TreeNode* cur, vector<int>& vec) {
        if (cur == NULL) return;
        traversal(cur->left, vec);
        vec.push_back(cur->val);
        traversal(cur->right, vec);
    }

    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> result;
        traversal(root, result);
        return result;
    }
};

四、367. 有效的完全平方数

367. 有效的完全平方数

Time / Space：时间4 ms击败25.44%，内存5.8 MB击败31.10%
Note：两个middle相乘会超过int，记着用long

class Solution {
public:
    bool isPerfectSquare(int num) {
        int leftIndex = 0;
        int rightIndex = num;

        while (leftIndex <= rightIndex) {
            long middle = leftIndex + ((rightIndex - leftIndex) / 2);
            if (middle * middle > num) {
                rightIndex = middle - 1;
            } else if (middle * middle < num) {
                leftIndex = middle + 1;
            } else {
                return true;
            }
        }
        return false;
    }
};

五、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;
    }
};

总结

因为二叉树题比较多，就分上下两篇记录了，系统的搞懂了前中后的两种遍历方式，下面就是广度优先遍历了，二叉树是很多算法的基础，非常重要！