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分析
- 找到一个就可以结束
- 二叉树还是递归用的多
- 深度优先遍历
/**
* 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 == nullptr){
return false;
}
if (root->left == nullptr and root->right == nullptr){
return sum == root->val;
}
bool b1=hasPathSum(root->left, sum - root->val);
if(b1){
return true;
}
bool b2=hasPathSum(root->right, sum - root->val);
if(b2){
return true;
}
return false;
}
};
感谢:https://github.com/glw0223/Leetcode-1/blob/master/src/0112-Path-Sum/0112.cpp