Leetcode 最小二叉树深度和删除数组重复元素

Number114 最小二叉树深度

class Solution {
public:
    int minDepth(TreeNode *root) {
        int depth=0;
        if(root==NULL)return 0; 
        int left=minDepth(root->left);
        int right=minDepth(root->right);
        if(root->left==NULL&&root->right!=NULL)
        {
            depth=right+1;
        }
        else if(root->right==NULL&&root->left!=NULL)
        {
            depth=left+1;
        }
        else if(root->left==NULL&&root->right==NULL)
        depth=depth+1;
        else
        depth=left<right ? left+1 : right+1;
        return depth;
    }
};

 Number 80 删除数组重复元素

class Solution {
public:
    int removeDuplicates(int A[], int n) {
       if(n==0)return 0;
       int i,j;
       int current=A[0];
       int count=1;
       for(i=1,j=1;i<n;i++)
        {
          if(A[i]==current)
          {
              count++;
              if(count<=2)
              A[j++]=current;     
          }
          if(A[i]!=current)
          {
             current=A[i];
             A[j++]=current;
             count=1;
          }
        }
        return j;
    }
};

Number 100 same tree

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSameTree(TreeNode *p, TreeNode *q) {
       if(p==NULL&&q==NULL)
       return true;
       if(p==NULL&&q!=NULL)return false;
       if(p!=NULL&&q==NULL)return false;
       if(p->val==q->val)
       {
         return isSameTree(p->left,q->left)&&isSameTree(p->right,q->right);
       } 
       else return false;        
    }
};

Number 94 二叉树中序遍历

class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
      vector<int> v;
      TreeNode *s[100];
      int top=-1;
      if(root==NULL)return v;
      TreeNode *p=root;
      //s[++top]=p;
      while(p||top!=-1)
      {
         while(p)
         {
            s[++top]=p;
            p=p->left;         
         } 
         if(top!=-1)
         {
            p=s[top];
            v.push_back(p->val);
            top--;
            p=p->right;                         
         } 
      }
      return v;              
    }
};