二叉树的深度，递归求
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 13 int maxDepth(TreeNode* root) { 14 int depth=0,high=0; 15 dfs(root,depth,high); 16 return depth; 17 } 18 void dfs(TreeNode* root,int &depth,int high){ 19 if(root==NULL) return; 20 high++; 21 if(high>depth) depth=high; 22 dfs(root->left,depth,high); 23 dfs(root->right,depth,high); 24 high--; 25 } 26 };
二叉搜索树的判断：
1 class Solution { 2 public: 3 bool isValidBST(TreeNode* root) { 4 bool res=true; 5 vector<int> vals; 6 dfs(root,vals); 7 for(int i=1;i<vals.size();i++){ 8 if(vals[i-1]>=vals[i]){ 9 res=false;break; 10 } 11 } 12 return res; 13 } 14 void dfs(TreeNode* root,vector<int> &vals){ 15 if(root==NULL) return; 16 dfs(root->left,vals); 17 vals.push_back(root->val); 18 dfs(root->right,vals); 19 } 20 };
对称二叉树：
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 isSymmetric(TreeNode* root) { 13 bool res=true; 14 vector<int> vals; 15 dfs(root,vals); 16 int i=0;int j=vals.size()-1; 17 while(i<j){ 18 if(vals[i]!=vals[j]){ 19 res=false;break; 20 } 21 i++;j--; 22 } 23 return res; 24 } 25 void dfs(TreeNode* root,vector<int> &vals){ 26 if(root==NULL) return; 27 dfs(root->left,vals); 28 vals.push_back(root->val); 29 dfs(root->right,vals); 30 } 31 };
二叉树层次遍历：
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 vector<vector<int>> levelOrder(TreeNode* root) { 13 vector<vector<int>> res; 14 if (root==NULL) return res; 15 queue<TreeNode*> q; 16 q.push(root); 17 while(!q.empty()){ 18 vector<int> level; 19 int len=q.size(); 20 for(int i=0;i<len;i++){ 21 TreeNode *front=q.front(); 22 level.push_back(front->val); 23 q.pop(); 24 if(front->left) q.push(front->left); 25 if(front->right) q.push(front->right); 26 } 27 res.push_back(level); 28 } 29 return res; 30 } 31 };