代码随想录|Day 14

Day 14 新年将至

一、理论学习

BFS 的使用场景总结：层序遍历、最短路径问题(https://leetcode.cn/problems/binary-tree-level-order-traversal/solutions/244853/bfs-de-shi-yong-chang-jing-zong-jie-ceng-xu-bian-l/)

BFS 的应用一：层序遍历

BFS 的应用二：最短路径

如果要更贴向代码二叉树的层序遍历可以看下面这个图

102二叉树的层序遍历

二、刷题部分

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) {
        vector<vector<int>>result;
        queue<TreeNode*>que;
        if(root!=nullptr) que.push(root);
        while(!que.empty()){
            int size=que.size();
            vector<int>res;
            while(size--){
                TreeNode* tmp=que.front();
                que.pop();
                res.push_back(tmp->val);
                if(tmp->left){
                    que.push(tmp->left);
                }
                if(tmp->right){
                    que.push(tmp->right);
                }
            }
            result.push_back(res);
        }
        return result;
    }
};

226. 翻转二叉树 - 力扣（LeetCode）

翻转二叉树

/**
 * 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);  // 中
        invertTree(root->left);         // 左
        invertTree(root->right);        // 右
        return root;

    }
};

101. 对称二叉树 - 力扣（LeetCode）

/**
 * 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 com(TreeNode* left,TreeNode* right){
        //递归终止条件
        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;
        //单层逻辑
        bool out=com(left->left,right->right);
        bool in=com(left->right,right->left);
        return in&&out;

    }
    bool isSymmetric(TreeNode* root) {
        return com(root->left,root->right);
       
        

    }
};