【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
110. Balanced Binary Tree
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as:
a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example 1:
Given the following tree
[3,9,20,null,null,15,7]
:
Return true.
Example 2:
Given the following tree
[1,2,2,3,3,null,null,4,4]
:
Return false.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
/*
对于每一个节点，我们通过
checkDepth
方法递归获得左右子树的深度，
如果子树是平衡的，则返回真实的深度，若不平衡，直接返回
-1
，
此方法时间复杂度
O(N)
，空间复杂度
O(H)
对于用递归求解的题，可以画递归树来分析
*/
class
Solution
{
public
:
bool
isBalanced
(
TreeNode
*
root
)
{
if
(
checkDepth
(
root
)
==
-
1
)
return
false
;
else
return
true
;
}
int
checkDepth
(
TreeNode
*
root
)
{
if
(!
root
)
return
0
;
//
为空时，返回深度
0
int
left
=
checkDepth
(
root
->
left
);
//
返回左子树的深度，如果不平衡，返回
-1
if
(
left
==
-
1
)
return
-
1
;
//只要哪里有-1一路返回，直到递归结束
int
right
=
checkDepth
(
root
->
right
);
//
返回右子树的深度
if
(
right
==
-
1
)
return
-
1
;
int
diff
=
abs
(
left
-
right
);
//
计算左右子树的深度差，如果不平衡，返回
-1
，平衡则返回实际深度
if
(
diff
>
1
)
return
-
1
;
else
return
1
+
max
(
left
,
right
);
}
};