題目:
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.
分析:
題目是要看給定的樹中是否存在一條從根到葉節點的路徑,該路徑上所有節點上值的和相加為給定的值。可以采用遞歸的思想,每次看該節點的右子樹節點的值是否為當前sum減去當前節點的值,左子樹也進行同樣的處理。
代碼:
class Solution {
public:
bool hasPathSum(TreeNode* root, int sum) {
if(!root)
return false;
if(!root->left&&!root->right)
{
if(root->val==sum)
return true;
else
return false;
}
return hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val);
}
};