【LeetCode 110_二叉树_遍历】Balanced Binary Tree

解法一：From top to bottom

 int treeHeight(TreeNode *T)
 {
     if (T == NULL)
         return ;

     return max(treeHeight(T->left), treeHeight(T->right)) + ;
 }

 bool isBalanced(TreeNode* root)
 {
     if (root == NULL)
         return true;

     int left_height = treeHeight(root->left);
     int right_height = treeHeight(root->right);
     if (abs(left_height - right_height) > )
         return false;

     if (!isBalanced(root->left) || !isBalanced(root->right))
         return false;

     return true;
 }

解法一递归判断子树是否为平衡二叉树的过程中，重复计算了多次子树的高度，时间复杂度大于O(N)。解法二将这个计算过程优化了，时间复杂度为O(N)。

解法二：From bottom to top

 int dfsTreeHeight(TreeNode *T)
 {
     if (T == NULL)
         return ;

     int left_height = dfsTreeHeight(T->left);
     if (left_height == -)
         return -;
     int right_height = dfsTreeHeight(T->right);
     if (right_height == -)
         return -;
     if (abs(left_height - right_height) > )
         return -;

     return max(left_height, right_height) + ;
 }

 bool isBalanced(TreeNode* root)
 {
     if (root == NULL)
         return true;

     return dfsTreeHeight(root) != -;
 }