代码随想录 Day15 LeetCode102. 二叉树的层序遍历

标签：right 层序 随想录 push qe 二叉树 front root left

102. 二叉树的层序遍历

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

107. 二叉树的层序遍历 II

class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>>result;
        queue<TreeNode*>qe;
        if(root!=NULL)qe.push(root);
        while(!qe.empty()){
            int size=qe.size();
            vector<int>temp;
            for(int i=0;i<size;i++){
                temp.push_back(qe.front()->val);
                if(qe.front()->left)qe.push(qe.front()->left);
                if(qe.front()->right)qe.push(qe.front()->right);
                qe.pop();
            }
            result.push_back(temp);
            
        }
        reverse(result.begin(),result.end());
        return result;
    }
};

199. 二叉树的右视图

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

226. 翻转二叉树

层序遍历做法

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

前序遍历（递归）做法

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

101. 对称二叉树

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        queue<TreeNode* >qe;
        if(root!=nullptr)qe.push(root);
        while(!qe.empty()){
            vector<int>result;
            int size=qe.size();
            for(int i=0;i<size;i++){
                if(qe.front()->left) qe.push(qe.front()->left),result.push_back(qe.front()->left->val);
                else result.push_back(101);
                if(qe.front()->right)qe.push(qe.front()->right),result.push_back(qe.front()->right->val);
                else result.push_back(101);
                qe.pop();
            }
            vector<int>result1(result);
            reverse(result.begin(),result.end());
            if(result1!=result)return 0;
        }
        return 1;
    }
};

后序遍历做法（递归）。

从根的左右子树进入，分为左右中，右左中两只遍历方式，分外侧和内侧比较

class Solution {
public:
    bool compare(TreeNode* left,TreeNode* right){
        if(left==NULL&&right==NULL)return 1;
        else if(left==NULL&&right!=NULL)return 0;
        else if(left!=NULL&&right==NULL)return 0;
        else if(left->val!=right->val)return 0;
        bool outside= compare(left->left,right->right);
        bool isside= compare(left->right,right->left);
        bool isSame= outside&&isside;
        return isSame;
    }
    bool isSymmetric(TreeNode* root) {
        if(root==NULL)return 1;
        return compare(root->left,root->right);
    }
};

 

标签：right,层序,随想录,push,qe,二叉树,front,root,left
From： https://www.cnblogs.com/zhishikele/p/17044196.html