Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree and sum = 22,
5
/
4 8
/ /
11 13 4
/
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
Analyse: DFS.
1. Recursion
Runtime: 12ms.
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool hasPathSum(TreeNode* root, int sum) { 13 if(!root) return false; 14 if(!root->left && !root->right) return sum == root->val; 15 16 return hasPathSum(root->left, sum - root->val) || 17 hasPathSum(root->right, sum - root->val); 18 } 19 };
2. Iteration-------need to think about a solution....
Runtime: