day12打卡

平衡二叉树

class Solution
{
public:
int getheight(TreeNode *root)
{
if(root == nullptr)
{
return 0;
}
int left = getheight(root->left);
int right = getheight(root->right);
return 1+max(left, right);
}
bool isBalanced(TreeNode *root)
{
if(root == nullptr)
{
return true;
}
bool l = isBalanced(root->left);
bool r = isBalanced(root->right);
int left = getheight(root->left);
int right = getheight(root->right);
if(abs(left-right) <= 1 && l && r)
{
return true;
}
else
{
return false;
}

}

};

二叉树所有的路径

没有想明白终止条件到空结点为啥不行
/**

• Definition for a binary tree node.
• 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:
  void traversal(TreeNode root, vector &path, vector &ret)
  {
  path.push_back(root->val);
  if(root->left == nullptr && root->right == nullptr)
  {
  string str;
  for(int i = 0; i < path.size() - 1; ++i)
  {
  str += to_string(path[i]);
  str += "->";
  }
  str += to_string(path[path.size()-1]);
  ret.push_back(str);
  }
  if(root->left)
  {
  traversal(root->left, path, ret);
  path.pop_back();
  }
  if(root->right)
  {
  traversal(root->right, path, ret);
  path.pop_back();
  }
  }
  vector binaryTreePaths(TreeNode root)
  {
  vector path;
  vector ret;
  if(root == nullptr)
  {
  return ret;
  }
  traversal(root, path, ret);
  return ret;
  }
  };

左叶子之和

class Solution
{
public:
int traversal(TreeNode *root)
{
int left = 0;
int right = 0;
if(root == nullptr)
{
return 0;
}
if(root->left == nullptr && root->right == nullptr)
{
return 0;
}
left = traversal(root->left);
if(root->left)
{
if(root->left->left == nullptr && root->left->right == nullptr)
{
left = root->left->val;
}

    }
    if(root->right)
    {
        right = traversal(root->right);
    }
    return left + right;
}
int sumOfLeftLeaves(TreeNode* root)
{
    if(root == nullptr)
    {
        return 0;
    }
    return traversal(root);
}

};

完全二叉树的结点个数
class Solution
{
public:
int countNodes(TreeNode *root)
{
if(root == nullptr)
{
return 0;
}
int left = countNodes(root->left);
int right = countNodes(root->right);
return 1+left+right;
}
};