解法一：From top to bottom
1 int treeHeight(TreeNode *T) 2 { 3 if (T == NULL) 4 return 0; 5 6 return max(treeHeight(T->left), treeHeight(T->right)) + 1; 7 } 8 9 bool isBalanced(TreeNode* root) 10 { 11 if (root == NULL) 12 return true; 13 14 int left_height = treeHeight(root->left); 15 int right_height = treeHeight(root->right); 16 if (abs(left_height - right_height) > 1) 17 return false; 18 19 if (!isBalanced(root->left) || !isBalanced(root->right)) 20 return false; 21 22 return true; 23 }
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
1 int dfsTreeHeight(TreeNode *T) 2 { 3 if (T == NULL) 4 return 0; 5 6 int left_height = dfsTreeHeight(T->left); 7 if (left_height == -1) 8 return -1; 9 int right_height = dfsTreeHeight(T->right); 10 if (right_height == -1) 11 return -1; 12 if (abs(left_height - right_height) > 1) 13 return -1; 14 15 return max(left_height, right_height) + 1; 16 } 17 18 bool isBalanced(TreeNode* root) 19 { 20 if (root == NULL) 21 return true; 22 23 return dfsTreeHeight(root) != -1; 24 }