二叉树相关题目
LeetCode 98. Validate Binary Search Tree(判断是否是二叉搜索树）
思路1：对树进行深度优先遍历，遍历到某个节点是判断该节点的左子树是否为二叉搜索树，右子树是否为二叉搜索树，左子树的最大值是否小于当前节点的值，右子树的最小值是否大于当前节点的值。
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool isValidBST(TreeNode* root) { 13 if(root == NULL) return true; 14 int maxv, minv; 15 return dfs(root, maxv, minv); 16 } 17 18 bool dfs(TreeNode* root, int &maxv, int &minv){ 19 maxv = minv = root->val; 20 if(root->left){ 21 int nowMaxv, nowMinv; 22 if(!dfs(root->left, nowMaxv, nowMinv)) 23 return false; 24 if(nowMaxv >= root->val) 25 return false; 26 maxv = max(maxv, nowMaxv); 27 minv = min(minv, nowMinv); 28 } 29 if(root->right){ 30 int nowMaxv, nowMinv; 31 if (!dfs(root->right, nowMaxv, nowMinv)) 32 return false; 33 if (nowMinv <= root->val) 34 return false; 35 maxv = max(maxv, nowMaxv); 36 minv = min(minv, nowMinv); 37 } 38 return true; 39 } 40 };
思路2：对树进行中序遍历，遍历的过程中判断当前值是否大于前一个值。
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool isValidBST(TreeNode* root) { 13 stack<TreeNode*> st; 14 int pre_val; 15 bool first_flag = true; 16 while(!st.empty() || root!=NULL){ 17 while(root!=NULL){ 18 st.push(root); 19 root = root->left; 20 } 21 root = st.top(); 22 st.pop(); 23 if(first_flag) 24 first_flag = false; 25 else if(root->val <= pre_val) 26 return false; 27 pre_val = root->val; 28 root = root->right; 29 } 30 return true; 31 } 32 };
2. LeetCode 101. Symmetric Tree(对称的二叉树）
递归实现
1 class Solution { 2 public: 3 bool isSymmetric(TreeNode* root) { 4 if(root == NULL) return true; 5 return judge(root->left, root->right); 6 } 7 8 bool judge(TreeNode* root1, TreeNode* root2){ 9 if(root1==NULL && root2==NULL) return true; 10 if(root1==NULL || root2==NULL) return false; 11 if(root1->val != root2->val) return false; 12 return judge(root1->left, root2->right) && 13 judge(root1->right, root2->left); 14 } 15 };
非递归实现
1 class Solution { 2 public: 3 bool isSymmetric(TreeNode* root) { 4 if(root==NULL) return true; 5 stack<TreeNode* > lstack, rstack; 6 TreeNode* lt = root->left, *rt = root->right; 7 while(lt || rt || lstack.size()){ 8 // 左右子树都不为空 9 while(lt && rt){ 10 lstack.push(lt), rstack.push(rt); 11 lt = lt->left, rt=rt->right; 12 } 13 // 一个为空，一个不为空 14 if(lt || rt) return false; 15 // 左右子树都为空 16 lt = lstack.top(), rt = rstack.top(); 17 lstack.pop(), rstack.pop(); 18 if(lt->val != rt->val) return false; 19 lt = lt->right, rt=rt->left; 20 } 21 return true; 22 } 23 };
3. LeetCode 297 Serialize and Deserialize Binary Tree(二叉树的序列化）
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Codec { 11 public: 12 13 // Encodes a tree to a single string. 14 string serialize(TreeNode* root) { 15 string ans = ""; 16 pre_order(root, ans); 17 return ans; 18 19 } 20 21 // Decodes your encoded data to tree. 22 TreeNode* deserialize(string data) { 23 int cur=0; 24 return decode(cur, data); 25 } 26 27 void pre_order(TreeNode* root, string& s){ 28 if(root == NULL){ 29 s += ","; 30 return; 31 } 32 s += to_string(root->val) + ","; 33 pre_order(root->left, s); 34 pre_order(root->right, s); 35 } 36 37 TreeNode* decode(int &cur, string &data){ 38 if(data[cur] == ','){ 39 cur++; 40 return NULL; 41 } 42 int next = data.find(',', cur); 43 int val = stoi(data.substr(cur, next-cur)); 44 TreeNode* node = new TreeNode(val); 45 cur = next + 1; 46 node->left = decode(cur, data); 47 node->right = decode(cur, data); 48 return node; 49 } 50 }; 51 52 // Your Codec object will be instantiated and called as such: 53 // Codec codec; 54 // codec.deserialize(codec.serialize(root));
