leecode 解题总结：101. Symmetric Tree

#include <iostream>#include <stdio.h>#include <vector>#include <queue>using namespace std;/*问题:Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).For example, this binary tree [1,2,2,3,4,4,3] is symmetric:    1   / \  2   2 / \ / \3  4 4  3But the following [1,2,2,null,3,null,3] is not:    1   / \  2   2   \   \   3    3Note:Bonus points if you could solve it both recursively and iteratively.分析:给定一颗二叉树，判定该二叉树是否是对称二叉树。所谓的对称二叉树应该是当前结点的左子树和右子树是对称的。设当前结点的左孩子为leftNode,右孩子为rightNode还是没有明白镜像如果用递归定义。leecode解法:https://leetcode.com/problems/symmetric-tree/?tab=Solutions仍然是基于结点的左右孩子是否相等的判定实际就是对(left , right)进行递归判断，如果left或者right中有一个非空，那么比较left是否等于right,不等返回false。如果left和right都不空，就需要对两者的值判定，如果值不等，肯定不等【最关键】:判定(left->left,right->right) 和(left->right , right->left)是否相等我之所以未能解答出来：尝试对(root)进行递归，而没有对两个结点递归输入:71 2 2 3 4 4 3输出:true关键:1 仍然是基于结点的左右孩子是否相等的判定实际就是对(left , right)进行递归判断，如果left或者right中有一个非空，那么比较left是否等于right,不等返回false。如果left和right都不空，就需要对两者的值判定，如果值不等，肯定不等【最关键】:判定(left->left,right->right) 和(left->right , right->left)是否相等我之所以未能解答出来：尝试对(root)进行递归，而没有对两个结点递归2注意树的问题如果不用递归，一般都需要用栈来解决。但是考虑到这是类似层序遍历的问题因此，用队列来做。//分别将自己的左右孩子压入到队列中//如果孩子为空就不压入，直接判断if(left->left){if(NULL == right->right){return false;}nodes1.push(left->left);nodes2.push(right->right);}//漏了一个判断else if(right->right){return false;}*/struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public://检查一个结点的左右孩子结点是否对称bool check(TreeNode* leftNode , TreeNode* rightNode){//牛逼，直接比较，因为如果一个为NULL就能直接比较if(NULL == leftNode || NULL == rightNode){return leftNode == rightNode;}if(leftNode->val != rightNode->val){return false;}//递归比较return check(leftNode->left , rightNode->right) && check(leftNode->right , rightNode->left);}    bool isSymmetric2(TreeNode* root) {if(!root){return true;}bool isOk = check(root->left , root->right);return isOk;    }    bool isSymmetric(TreeNode* root) {if(!root){return true;}queue<TreeNode*> nodes1;queue<TreeNode*> nodes2;if(root->left){if(NULL == root->right){return false;}nodes1.push(root->left);nodes2.push(root->right);}//如果左孩子为空，右孩子不空，返回falseelse if(root->right){return false;}TreeNode* left;TreeNode* right;//确保队列里面任意两个结点都是非空的，便于进行比较；空的部分，直接提前比较while((!nodes1.empty()) && !(nodes2.empty())){left = nodes1.front();right = nodes2.front();nodes1.pop();nodes2.pop();if(left->val != right->val){return false;}//分别将自己的左右孩子压入到队列中//如果孩子为空就不压入，直接判断if(left->left){if(NULL == right->right){return false;}nodes1.push(left->left);nodes2.push(right->right);}//漏了一个判断else if(right->right){return false;}if(left->right){if(NULL == right->left){return false;}nodes1.push(left->right);nodes2.push(right->left);}else if(right->left){return false;}}//如果其中某个队列还有元素剩余，不行if(nodes1.empty() && nodes2.empty()){return true;}else{return false;}    }};//采用递归的方式构建二叉树,确定根节点，需要后面明确左孩子和右孩子，当前结点为i,左孩子为2*i,右孩子为2*i+1，默认i从1开始TreeNode* buildTree(vector<int>& nums , int pos ,int len , vector<TreeNode*>& nodes){if(nums.empty()){return NULL;}//判断是否遍历结束if(pos > len){if(!nodes.empty()){return nodes.at(0);}else{return NULL;}}//构建根节点if(nodes.empty()){TreeNode* root = new TreeNode(nums.at(0));nodes.push_back(root);}//取出当前结点TreeNode* current = NULL;if(0 <= pos - 1 && pos - 1 <= nodes.size() - 1){current = nodes.at(pos - 1);}//构建当前结点的左右孩子结点,int j = 2 * pos;if(j <= len){TreeNode* left = new TreeNode(nums.at(j-1));nodes.push_back(left);if(current){current->left = left;}}if(j + 1 <= len){TreeNode* right = new TreeNode(nums.at(j));nodes.push_back(right);if(current){current->right = right;}}buildTree(nums , pos + 1 , len , nodes);return nodes.at(0);}void deleteTree(TreeNode* root){if(!root){return;}if(NULL == root->left && NULL == root->right){delete root;root = NULL;}if(root){deleteTree(root->left);deleteTree(root->right);}}void print(vector<int>& result){if(result.empty()){cout << "no result" << endl;return;}int size = result.size();for(int i = 0 ; i < size ; i++){cout << result.at(i) << " " ;}cout << endl;}void process(){ vector<int> nums; int value; int num; Solution solution; vector<TreeNode*> nodes; while(cin >> num ) { nums.clear(); nodes.clear(); for(int i = 0 ; i < num ; i++) { cin >> value; nums.push_back(value); } //初始pos从1开始 TreeNode* root = buildTree(nums , 1, nums.size() , nodes); bool isSym = solution.isSymmetric(root); if(isSym) { cout << "true" << endl; } else { cout << "false" << endl; } //删除二叉树 deleteTree(root); }}int main(int argc , char* argv[]){process();getchar();return 0;}