112. Path Sum
2017-06-19 本文已影响0人
YellowLayne
1.描述
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.
2.分析
3.代码
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
bool hasPathSum(struct TreeNode* root, int sum) {
if (NULL == root) return false;
if (NULL == root->left && NULL == root->right) return root->val == sum;
if (NULL == root->left) return hasPathSum(root->right, sum - root->val);
if (NULL == root->right) return hasPathSum(root->left ,sum - root->val);
bool left = hasPathSum(root->left, sum - root->val);
bool right = hasPathSum(root->right, sum - root->val);
return left || right;
}