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222. Count Complete Tree Nodes

Solution - Recursion
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: int countNodes(TreeNode* root) { int cnt = 0; if (!root) return 0; return 1 + countNodes(root->left) + countNodes(root->right); } };
  • T:
    O(n)
  • S:
    O(n)
Solution 2
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: int countNodes(TreeNode* root) { int leftHeight = 0, rightHeight = 0; TreeNode* leftNode = root; TreeNode* rightNode = root; while (leftNode) { ++leftHeight; leftNode = leftNode->left; } while (rightNode) { ++rightHeight; rightNode = rightNode->right; } if (leftHeight == rightHeight) { return pow(2, leftHeight) - 1; } return countNodes(root->left) + countNodes(root->right) + 1; } };
  • T:
    O(lognlogn)
  • S:
    O(n)