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二叉树经典面试题：：
目录
二叉树经典面试题：：
1.根据二叉树创建字符串
2.二叉树的层序遍历
3.二叉树的层序遍历II
4.二叉树的最近公共祖先
5.二叉树与双向链表
6.从前序与中序序列构造二叉树
7.从中序与后序序列构造二叉树
8.非递归实现二叉树的前序遍历
9.非递归实现二叉树的中序遍历
10.非递归实现二叉树的后序遍历
11.前K个高频单词
12.在长度2N的数组中找出重复N次的元素
13.两个数组的交集I
14.两个数组的交集II
15.两句话中的不常见单词
1.根据二叉树创建字符串
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:string tree2str(TreeNode* root) {if (root == nullptr){return string();}string str;str += to_string(root->val);if (root->left != nullptr){str += '(';str += tree2str(root->left);str += ')';}else if(root->left == nullptr && root->right != nullptr){str += "()";}if (root->right != nullptr){str += '(';str += tree2str(root->right);str += ')';}//右边为空则不处理return str;}
};2.二叉树的层序遍历
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*> q;vector<vector<int>> vv;int levelSize = 0;if (root != nullptr){q.push(root);levelSize = 1;}while (!q.empty()){//一层一层出vector<int> levelv;while (levelSize--){TreeNode* front = q.front();q.pop();levelv.push_back(front->val);if (front->left != nullptr){q.push(front->left);}if (front->right != nullptr){q.push(front->right);}}vv.push_back(levelv);//当前一层出完 下一层全部进入队列 那么size就是下一层的数据个数levelSize = q.size();}return vv;}
};3.二叉树的层序遍历II
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*> q;vector<vector<int>> vv;int levelSize = 0;if (root != nullptr){q.push(root);levelSize = 1;}while (!q.empty()){vector<int> levelv;while (levelSize--){TreeNode* front = q.front();q.pop();levelv.push_back(front->val);if (front->left != nullptr){q.push(front->left);}if (front->right != nullptr){q.push(front->right);}}vv.push_back(levelv);levelSize = q.size();}reverse(vv.begin(), vv.end());return vv;}
};4.二叉树的最近公共祖先
struct TreeNode{int val;TreeNode *left;TreeNode *right;TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution
{
public:bool GetPath(TreeNode* root, TreeNode* x, stack<TreeNode*>& path){if (root == nullptr){return false;}path.push(root);if (root == x){return true;}if (GetPath(root->left, x, path)){return true;}if (GetPath(root->right, x, path)){return true;}path.pop();return false;}TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q){stack<TreeNode*> pPath;stack<TreeNode*> qPath;GetPath(root, p, pPath);GetPath(root, q, qPath);//找路径相遇点while (pPath.size() != qPath.size()){if (pPath.size() > qPath.size()){pPath.pop();}else{qPath.pop();}}while (pPath.top() != qPath.top()){pPath.pop();qPath.pop();}return pPath.top();}};5.二叉树与双向链表
struct TreeNode
{int val;struct TreeNode *left;struct TreeNode *right;TreeNode(int x) :val(x), left(NULL), right(NULL) {}
};
class Solution
{
public:void InOrderConvert(TreeNode* cur, TreeNode*& prev){if (cur == nullptr){return;}InOrderConvert(cur->left, prev);cur->left = prev;if (prev != nullptr){prev->right = cur;}prev = cur;InOrderConvert(cur->right, prev);}TreeNode* Convert(TreeNode* pRootOfTree) {TreeNode* prev = nullptr;InOrderConvert(pRootOfTree, prev);TreeNode* head = pRootOfTree;while (head != nullptr && head->left != nullptr){head = head->left;}return head;}
};6.从前序与中序序列构造二叉树
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* _buildTree(vector<int>& preorder, vector<int>& inorder, int& prei, int inbegin, int inend) {//中序区间不存在则是空树if (inbegin > inend){return nullptr;}//划分左右区间int rooti = inbegin;while (rooti <= inend){if (inorder[rooti] ==preorder[prei]){break;}++rooti;}TreeNode* root = new TreeNode(preorder[prei]);++prei;//[inbegin,rooti-1] rooti [rooti+1,inend]root->left = _buildTree(preorder, inorder, prei, inbegin, rooti - 1);root->right = _buildTree(preorder, inorder, prei, rooti + 1, inend);return root;}TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder){int pi = 0;return _buildTree(preorder, inorder, pi, 0, inorder.size() - 1);}
};7.从中序与后序序列构造二叉树
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* _buildTree(vector<int>& inorder, int inbegin, int inend, vector<int>& postorder, int& pi){if (inbegin > inend){return nullptr;}TreeNode* root = new TreeNode(postorder[pi]);pi--;//将该子树的中序遍历序列划分为其左子树和右子树的中序遍历序列int rooti = inbegin;while (inorder[rooti] != root->val){rooti++;}root->right = _buildTree(inorder, rooti + 1, inend, postorder, pi);root->left = _buildTree(inorder, inbegin, rooti - 1, postorder, pi);return root;}TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {int i = postorder.size() - 1; return _buildTree(inorder, 0, inorder.size() - 1, postorder, i);}
};8.非递归实现二叉树的前序遍历
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> preorderTraversal(TreeNode* root) {vector<int> v;stack<TreeNode*> st;TreeNode* cur = root;while (!st.empty() || cur != nullptr){//准备开始访问一棵树//访问左路结点//左路结点的右子树while (cur != nullptr){v.push_back(cur->val);st.push(cur);cur = cur->left;}//左路结点右子树TreeNode* top = st.top();st.pop();//循环子问题访问右子树cur = top->right;}return v;}
};9.非递归实现二叉树的中序遍历
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> inorderTraversal(TreeNode* root) {vector<int> v;stack<TreeNode*> st;TreeNode* cur = root;while (cur || !st.empty()){while (cur){st.push(cur);cur = cur->left;}TreeNode* top = st.top();st.pop();v.push_back(top->val);//访问右子树cur = top->right;}return v;}
};10.非递归实现二叉树的后序遍历
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> postorderTraversal(TreeNode* root){vector<int> v;stack<TreeNode*> st;TreeNode* cur = root;TreeNode* prev = nullptr;while (cur || !st.empty()){while (cur){st.push(cur);cur = cur->left;}TreeNode* top = st.top();//1.top的右子树为空或者右子树已经访问过了(上一个访问的结点是右子树的根)//那么说明右子树不用访问或者访问过了，可以访问根top//2.右子树不为空且没有访问，则迭代子问题访问if (top->right == nullptr || top->right == prev){st.pop();v.push_back(top->val);prev = top;}else{cur = top->right;}}return v;}
};
11.前K个高频单词
class Solution
{
public:struct Compare{bool operator()(const pair<int, string>& l, const pair<int, string>& r){return l.first > r.first;}};
public:vector<string> topKFrequent(vector<string>& words, int k){map<string, int> countMap;for (auto& str : words){countMap[str]++;}vector<pair<int, string>> v;for (auto& kv : countMap){v.push_back(make_pair(kv.second, kv.first));cout << kv.first << ":" << kv.second << endl;}stable_sort(v.begin(), v.end(), Compare());vector<string> ret;for (int i = 0; i < k; i++){ret.push_back(v[i].second);}return ret;}
};12.在长度2N的数组中找出重复N次的元素
class Solution
{
public:int repeatedNTimes(vector<int>& A) {size_t N = A.size() / 2;// 用unordered_map统计每个元素出现的次数unordered_map<int, int> m;for (auto e : A){m[e]++;}// 找出出现次数为N的元素for (auto& e : m){if (e.second == N){return e.first;}}return -1;}
};
13.两个数组的交集I
class Solution
{
public:vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {// 用unordered_set对nums1中的元素去重unordered_set<int> s1;for (auto e : nums1){s1.insert(e);}// 用unordered_set对nums2中的元素去重unordered_set<int> s2;for (auto e : nums2){s2.insert(e);}// 遍历s1，如果s1中某个元素在s2中出现过，即为交集vector<int> vRet;for (auto e : s1){if (s2.find(e) != s2.end()){vRet.push_back(e);}}return vRet;}
};
14.两个数组的交集II
//哈希表
class Solution
{
public:vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {//为了降低空间复杂度，应该先将元素个数较少的数组放入哈希表if (nums1.size() > nums2.size()){return intersect(nums2, nums1);}//1、将nums1中的每个数字以及对应出现的次数放入哈希表中unordered_map<int, int> um;for (auto e : nums1){um[e]++;}//2、遍历nums2，找出nums1和nums2的交集vector<int> vRet;for (auto e : nums2){if (um.count(e)) //该元素在哈希表中{vRet.push_back(e); //该元素属于交集um[e]--; //减少该元素在哈希表中出现的次数if (um[e] == 0) //该元素的次数已经变为0了{um.erase(e); //将该元素从哈希表中删除}}}return vRet;}
};
15.两句话中的不常见单词
class Solution
{
public:vector<string> uncommonFromSentences(string s1, string s2) {unordered_map<string, int> um;//1、将两个字符串拼接起来，中间加一个" "字符串string str = s1 + " " + s2;//2、将拼接后字符串当中的单词放入哈希表size_t start = 0; //字头size_t pos = str.find(' '); //字尾while (pos != string::npos) //直到找到字符串结尾{string word = str.substr(start, pos - start); //将单词提取出来um[word]++; //将单词放入哈希表start = pos + 1; //更新字头pos = str.find(' ', start); //更新字尾}//将最后一个单词放入哈希表string word = str.substr(start);um[word]++;//3、找出哈希表中只出现过一次的单词，即不常见单词vector<string> vRet;for (auto e : um){if (e.second == 1) //只出现过一次的单词{vRet.push_back(e.first);}}return vRet;}
};