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.
Note: A leaf is a node with no children.
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.
递归方法：
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 private: 12 bool isLeaf(TreeNode *root) { 13 return (root->left == NULL) && (root->right == NULL); 14 } 15 public: 16 bool hasPathSum(TreeNode* root, int sum) { 17 if (root == NULL) 18 return false; 19 if (isLeaf(root) && sum == root->val) 20 return true; 21 return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val); 22 } 23 };
非递归：
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 == NULL) 14 return false; 15 queue<TreeNode*> q; 16 q.push(root); 17 TreeNode *temp; 18 while (!q.empty()) { 19 int size = q.size(); 20 while (size--) { 21 temp = q.front(); 22 q.pop(); 23 if (temp->left != NULL) { 24 q.push(temp->left); 25 temp->left->val += temp->val; 26 } 27 if (temp->right != NULL) { 28 q.push(temp->right); 29 temp->right->val += temp->val; 30 } 31 if (temp->left == NULL && temp->right == NULL && temp->val == sum) { 32 return true; 33 } 34 35 } 36 } 37 return false; 38 } 39 };