当前位置： 面试刷题>> 二叉树的路径和Ⅰ (经典算法题500道)
### 题目描述补充
**题目：二叉树的路径和Ⅰ**
给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，这条路径上所有节点值相加等于目标和。
**说明**:
- 叶子节点是指没有子节点的节点。
- 路径可以从任意节点开始，不一定从根节点开始。
**示例**:
给定如下二叉树和目标和 `sum = 22`，
```
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
```
返回 `true`，因为存在一条路径 `5->4->11->2`，路径和等于 22。
### PHP 示例代码
```php
val = $val;
$this->left = $left;
$this->right = $right;
}
}
function hasPathSum($root, $sum) {
if ($root === null) {
return false;
}
// 如果当前节点是叶子节点且值等于目标和，则返回true
if ($root->left === null && $root->right === null && $root->val === $sum) {
return true;
}
// 递归检查左子树和右子树
return hasPathSum($root->left, $sum - $root->val) || hasPathSum($root->right, $sum - $root->val);
}
// 示例用法
$root = new TreeNode(5);
$root->left = new TreeNode(4);
$root->right = new TreeNode(8);
$root->left->left = new TreeNode(11);
$root->left->left->left = new TreeNode(7);
$root->left->left->right = new TreeNode(2);
$root->right->left = new TreeNode(13);
$root->right->right = new TreeNode(4);
$root->right->right->left = new TreeNode(5);
$root->right->right->right = new TreeNode(1);
$sum = 22;
$result = hasPathSum($root, $sum);
echo $result ? "true" : "false"; // 输出 true
?>
```
### Python 示例代码
```python
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def hasPathSum(root, sum):
if not root:
return False
if not root.left and not root.right and root.val == sum:
return True
return hasPathSum(root.left, sum - root.val) or hasPathSum(root.right, sum - root.val)
# 示例用法
root = TreeNode(5)
root.left = TreeNode(4)
root.right = TreeNode(8)
root.left.left = TreeNode(11)
root.left.left.left = TreeNode(7)
root.left.left.right = TreeNode(2)
root.right.left = TreeNode(13)
root.right.right = TreeNode(4)
root.right.right.left = TreeNode(5)
root.right.right.right = TreeNode(1)
sum = 22
result = hasPathSum(root, sum)
print(result) # 输出 True
```
### JavaScript 示例代码
```javascript
function TreeNode(val, left, right) {
this.val = (val===undefined ? 0 : val)
this.left = (left===undefined ? null : left)
this.right = (right===undefined ? null : right)
}
function hasPathSum(root, 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);
}
// 示例用法
let root = new TreeNode(5);
root.left = new TreeNode(4);
root.right = new TreeNode(8);
root.left.left = new TreeNode(11);
root.left.left.left = new TreeNode(7);
root.left.left.right = new TreeNode(2);
root.right.left = new TreeNode(13);
root.right.right = new TreeNode(4);
root.right.right.left = new TreeNode(5);
root.right.right.right = new TreeNode(1);
let sum = 22;
let result = hasPathSum(root, sum);
console.log(result); // 输出 true
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```