二叉树面试题

二叉树的定义

struct TreeNode{    int val;    TreeNode *left,*right;    TreeNode(int x):val(x),left(0),right(0){}};

1.求二叉树的结点个数

int f(TreeNode *root){    if(!root)return 0;    return f(root->left)+f(root->right)+1;}

2.求二叉树的深度

int f(TreeNode *root){    if(!root)return 0;    return max(f(root->left),f(root->right))+1;}

3.前、中、后序遍历

void preorderTraversal(TreeNode *root)  {      if(root)      {          cout<<root->val<<' ';        preorderTraversal(root->left);          preorderTraversal(root->right);       }   }  void inorderTraversal(TreeNode *root)  {      if(root)      {          inorderTraversal(root->left);        cout<<root->val<<' ';        inorderTraversal(root->right);       }   }void postorderTraversal(TreeNode *root){    if(root)    {        postorderTraversal(root->left);        postorderTraversal(root->right);        cout<<root->val<<' ';    }}

http://blog.csdn.net/hz5034/article/details/45582493

http://blog.csdn.net/hz5034/article/details/45599957

4.层序遍历

void levelorderTraverse(TreeNode *root)  {if(!root)return;   queue<TreeNode *> q;q.push(root);TreeNode *p;while(!q.empty())  {  p=q.front();  q.pop();  cout<<p->val<<' ';  if(!p->left)q.push(p->left);  if(!p->right)q.push(p->right);  }  }

http://blog.csdn.net/hz5034/article/details/44275239

5.重建二叉树

//前序+中序TreeNode *rebuild(int pre[],int in[],int n){if(n==0)return 0;TreeNode *root=new TreeNode(pre[0]);int i;for(i=0;i<n;++i){if(in[i]==root->val)break;}//重建左子树root->left=rebuild(pre+1,in,i);//重建右子树root->right=rebuild(pre+i+1,in+i+1,n-i-1);return root;}//中序+后序TreeNode *rebuild(int in[],int post[],int n){if(n==0)return 0;TreeNode *root=new TreeNode(post[n-1]);int i;for(i=0;i<n;++i){if(in[i]==root->val)break;}//重建左子树root->left=rebuild(in,post,i);//重建右子树root->right=rebuild(in+i+1,post+i,n-i-1);   return root;}

6.求二叉树中节点的最大距离

struct result{int maxDistance;int maxDepth;};result f(TreeNode *root){if(!root){result r={0,-1};return r;}result left=f(root->left);result right=f(root->right);result r;r.maxDistance=max(max(left.maxDistance,right.maxDistance),left.maxDepth+right.maxDepth+1);r.maxDepth=max(left.maxDepth,right.maxDepth)+1;return r;}

http://www.cnblogs.com/miloyip/archive/2010/02/25/1673114.html

7.二叉树中和为某一值的路径

深搜一：

class Solution {public:    vector<vector<int>> pathSum(TreeNode* root, int sum) {vector<vector<int> > result;vector<int> path;dfs(root,sum,result,path);return result;}void dfs(TreeNode *root,int sum,vector<vector<int> > &result,vector<int> &path){if(!root)return;path.push_back(root->val);if(!root->left&&!root->right&&sum==root->val)result.push_back(path);dfs(root->left,sum-root->val,result,path);dfs(root->right,sum-root->val,result,path);path.pop_back();}};

深搜二：

class Solution {public:        vector<vector<int>> pathSum(TreeNode* root, int sum) {vector<vector<int> > result;vector<int> path;int s=0;dfs(root,sum,result,path,s);return result;}void dfs(TreeNode *root,int sum,vector<vector<int> > &result,vector<int> &path,int &s)    {if(!root)return;path.push_back(root->val);s+=root->val;if(!root->left&&!root->right&&s==sum)result.push_back(path);dfs(root->left,sum,result,path,s);dfs(root->right,sum,result,path,s);s-=path.back();path.pop_back();}};