70. Climbing Stairs
This problem is a Fibonacci problem.
F(n)=F(n-1)+F(n-2);
Solving this problem by recursion ,we will do a lot of same recursion.
Example:
F(10)=F(9)+F(8);
F(9)=F(8)+F(7);
we calculate F(8) twice,when n is large,this will increase as a rate of n's exponent.
So a more efficient way to solve this problem is from Bottom to Top.
Calculate F(0) ,F(1);
then F(2).........
//F(n) = F(n-1) + F(n-2) class Solution { public: int climbStairs(int n) { //if(n<=2) return n; //0,1 int f0 = 0, f1=1, steps =0; for(int i=0; i<n; i++){ steps = f0+f1; f0 = f1; f1 = steps; } return steps; } }; //Recursion time out /*class Solution { public: int climbStairs(int n) { if(n<=2) return n; //0,1 int steps = 0; steps += climbStairs(n-1); steps += climbStairs(n-2); return steps; } };*/
LeetCode -- 1038. Binary Search Tree to Greater Sum Tree
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int preroot = 0; TreeNode* bstToGst(TreeNode* root) { if(root->right) bstToGst(root->right); preroot = root->val = root->val + preroot; if(root->left) bstToGst(root->left); return root; } };