题目
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,
return true, as there exist a root-to-leaf path 5->4->11->2
which sum is 22.
分析
判断给定树中有无 【跟—>叶子】路径节点值之和为 给定sum
递归实现思想:
- 空树,返回false
- 单根节点,判断,若等于sum返回true,否则返回false
- 更新sum -= root—>val 递归判断左右子树,其中一真即真。
AC代码
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool hasPathSum(TreeNode* root, int sum) {
//空树返回false
if (root == NULL)
{
return false;
}
//若是叶子节点,判断返回
else if (!root->left && !root->right)
{
if (root->val == sum)
return true;
else
return false;
}
else
{
//否则,递归判断左右子树
sum -= root->val;
return hasPathSum(root->left, sum) || hasPathSum(root->right, sum);
}
}
};