LeetCode 110. Balanced Binary Tree

本博客参考自：http://www.cnblogs.com/grandyang/p/4045660.html
Solution1：双递归解决问题！
高度平衡二叉树是每一个节点的两个字数的深度差不能超过1，那么我们肯定需要一个求各个点深度的函数，然后对每个节点的两个子树来比较深度差，时间复杂度为O(NlgN)，代码如下：

class Solution {
public:bool isBalanced(TreeNode *root) {if (!root) return true;if (abs(getDepth(root->left) - getDepth(root->right)) > 1) return false;return isBalanced(root->left) && isBalanced(root->right);    }int getDepth(TreeNode *root) {if (!root) return 0;return 1 + max(getDepth(root->left), getDepth(root->right));}
};

Solution2：
上面那个方法正确但不是很高效，因为每一个点都会被上面的点计算深度时访问一次，我们可以进行优化。方法是如果我们发现子树不平衡，则不计算具体的深度，而是直接返回-1。那么优化后的方法为：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)，参见代码如下：

class Solution {
public:    bool isBalanced(TreeNode *root) {if (checkDepth(root) == -1) return false;else return true;}int checkDepth(TreeNode *root) {if (!root) return 0;int left = checkDepth(root->left);if (left == -1) return -1;int right = checkDepth(root->right);if (right == -1) return -1;int diff = abs(left - right);if (diff > 1) return -1;else return 1 + max(left, right);}
};

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