问题:路径总和
解题思路
采用递归的思路
- root为NULL返回false
- root为叶节点时判断节点值和sum是否相等
- root有孩子时,对有孩子的分支进行判断对结果进行逻辑或运算
C++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) {
if(!root) return false;//NULL返回false
if(!root->left && !root->right) return root->val == sum;//root是根节点
//root有孩子节点
bool flag = false;
if(root->left) flag = flag || hasPathSum(root->left, sum - root->val);
if(root->right) flag = flag || hasPathSum(root->right, sum - root->val);
return flag;
}
};