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.
Note: A leaf is a node with no children.
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.
给定二叉树和求和,确定树是否具有根到叶路径,使得沿路径的所有值相加等于给定的总和。
思路
错误分析
刚开始我的代码是这个样子滴
1.如果root为NULL且sum为0,那么就相当于找到路径
2.不然就将目标sum减去已找到的路径值,递归遍历左右子树中有没有路径
未通过的原因
Input:[]
0
Output:true
Expected:false
也就是说理解错题意了,当没有根结点的时候,哪怕是sum为0都找不到路径的
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
bool hasPathSum(struct TreeNode* root, int sum) {
if(!root && sum == 0) return true;
if(root){
sum -= root->val;
return hasPathSum(root->left, sum) || hasPathSum(root->right, sum);
}
return false;
}
矫正
错在哪里了呢?
我们把返回true的条件重新梳理一下,当root为NULL的时候是不能返回true的
所以新的判断条件就变成:root && root->left == NULL && root->right == NULL && sum == root->val
注意这里sum不是判断== 0,而是root->val
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
bool hasPathSum(struct TreeNode* root, int sum) {
if(root && root->left == NULL && root->right == NULL && sum == root->val) return true;
if(root){
sum -= root->val;
return hasPathSum(root->left, sum) || hasPathSum(root->right, sum);
}
return false;
}