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.
递归,DFS,用stack实现还没想好怎么弄。。
1 /** 2 * Definition for binary tree 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 14 if(root == NULL) 15 return false; 16 if(root->left == NULL && root->right == NULL) 17 { 18 if(root->val == sum) 19 return true; 20 else 21 return false; 22 } 23 bool left = false; 24 bool right= false; 25 26 if(root->left) 27 left = hasPathSum(root->left, sum - root->val); 28 if(root->right) 29 right = hasPathSum(root->right, sum - root->val); 30 31 return (left || right); 32 } 33 };