Title: Enter a binary tree, the binary tree is determined whether a balanced binary tree.

Solution a:

ideas: the use of a function of the depth questions binary tree, each node of the comparison to a depth of about. However, this method requires repeated a plurality of times to traverse the nodes, such that the time efficiency is not high.

```
class Solution {
public:
bool IsBalanced_Solution(TreeNode* pRoot) {
if(pRoot == nullptr)
return true;
int left = TreeDepth(pRoot->left);
int right =TreeDepth(pRoot->right);
int diff = abs(left-right);
if (diff >1)
return false;
return IsBalanced_Solution(pRoot->left)&&IsBalanced_Solution(pRoot->right);
}
int TreeDepth(TreeNode* pRoot)
{
if(pRoot == nullptr)
return 0;
return 1+max(TreeDepth(pRoot->left),TreeDepth(pRoot->right));
}
};
```

Solution two:

the mind: if the subtree is not a balanced binary tree, a depth directly returns -1

```
class Solution {
public:
bool IsBalanced_Solution(TreeNode* pRoot) {
if (pRoot==nullptr)
return true;
return TreeDeep(pRoot)!=-1;
}
int TreeDeep(TreeNode* pRoot)
{
if (pRoot==nullptr)
return 0;
int left = TreeDeep(pRoot->left);
if (left==-1)
return -1;
int right =TreeDeep(pRoot->right);
if (right==-1)
return -1;
int diff = abs(left-right);
return diff>1?-1:max(1+left,1+right);
}
};
```