Balanced binary tree judgment of Li Kou 110 -- classic problems

Title Overview ：

Ideas ： Beyond all doubt , Most of the basic topics about binary trees must use the idea of recursion . This topic is no exception .
Here we use the idea of postorder traversal , First traverse the left subtree , And then I go through the right subtree , Final judgment root . Every time you traverse , We all have to find the depth of the root node corresponding to the left and right subtrees , Then make a comparison , The absolute value is less than 2 Then this node meets , Traverse in turn , It is not a balanced tree until all nodes are satisfied .

In this code, we must first write a code to find the depth of the binary tree , It's very simple. I won't go into details here .

The code is as follows :

int maxDepth(struct TreeNode* root)
{
    
    if(root==NULL)
    {
    
        return 0;
    }
    int leftdepth=maxDepth(root->left);
    int rightdepth=maxDepth(root->right);
    return leftdepth>rightdepth?leftdepth+1:rightdepth+1;
}

bool isBalanced(struct TreeNode* root)
{
    
    if(root==NULL)
    {
    
        return true;
    }
    return abs(maxDepth(root->left)-maxDepth(root->right))<2&&
    isBalanced(root->left)&&isBalanced(root->right);    // This expression means that each node must return true Finally, I came back true

}