solution.md

# 路径总和

<p>给你二叉树的根节点 <code>root</code> 和一个表示目标和的整数 <code>targetSum</code> ，判断该树中是否存在 <strong>根节点到叶子节点</strong> 的路径，这条路径上所有节点值相加等于目标和 <code>targetSum</code> 。</p>

<p><strong>叶子节点</strong> 是指没有子节点的节点。</p>

<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsum1.jpg" style="width: 500px; height: 356px;" />
<pre>
<strong>输入：</strong>root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
<strong>输出：</strong>true
</pre>

<p><strong>示例 2：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsum2.jpg" />
<pre>
<strong>输入：</strong>root = [1,2,3], targetSum = 5
<strong>输出：</strong>false
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>root = [1,2], targetSum = 0
<strong>输出：</strong>false
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>树中节点的数目在范围 <code>[0, 5000]</code> 内</li>
	<li><code>-1000 <= Node.val <= 1000</code></li>
	<li><code>-1000 <= targetSum <= 1000</code></li>
</ul>

<p>以下<span style="color:red">错误</span>的选项是？</p>

## aop

### before

```c
#include <bits/stdc++.h>
using namespace std;

struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode() : val(0), left(nullptr), right(nullptr) {}
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
```

### after

```c

```

## 答案

```c
其他三个都是错的
```
## 选项

### A

```c
class Solution
{
public:
    bool hasPathSum(TreeNode *root, int sum)
    {
        if (root == NULL)
        {
            return false;
        }
        vector<TreeNode *> node_stack;
        vector<int> sum_stack;
        node_stack.push_back(root);
        sum_stack.push_back(sum - root->val);

        TreeNode *node;
        int curr_sum;
        while (!node_stack.empty())
        {
            node = node_stack.back();
            node_stack.pop_back();
            curr_sum = sum_stack.back();
            sum_stack.pop_back();
            if ((node->right == NULL) && (node->left == NULL) && (curr_sum == 0))
            {
                return true;
            }
            if (node->right != NULL)
            {
                node_stack.push_back(node->right);
                sum_stack.push_back(curr_sum - node->right->val);
            }
            if (node->left != NULL)
            {
                node_stack.push_back(node->left);
                sum_stack.push_back(curr_sum - node->left->val);
            }
        }
        return false;
    }
};
```

### B

```c

class Solution
{
public:
    bool hasPathSum(TreeNode *root, int sum)
    {
        if (root == NULL)
            return false;
        stack<pair<TreeNode *, int>> s;
        s.push(make_pair(root, sum));
        TreeNode *node = root;
        int currSum;
        while (!s.empty())
        {
            node = s.top().first;
            currSum = s.top().second;
            s.pop();
            if (node->left == NULL && node->right == NULL && node->val == currSum)
                return true;
            if (node->right)
                s.push(make_pair(node->right, currSum - node->val));
            if (node->left)
                s.push(make_pair(node->left, currSum - node->val));
        }
        return false;
    }
};
```

### C

```c
class Solution
{
public:
    bool hasPathSum(TreeNode *root, int sum)
    {
        if (!root)
            return false;
        if (!root->left && !root->right && root->val == sum)
            return true;
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
    }
};
```