当前位置： 面试刷题>> 路径总和(经典算法150题)
### 题目描述
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。
**注意**：叶子节点是指没有子节点的节点。
### 示例
给定如下二叉树和目标和 `sum = 22`：
```
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
```
返回 `true`，因为存在一条路径 `5->4->11->2`，其和为 22。
### PHP 示例代码
```php
val = $val;
$this->left = $left;
$this->right = $right;
}
}
class Solution {
/**
* @param TreeNode $root
* @param Integer $sum
* @return Boolean
*/
function hasPathSum($root, $sum) {
if ($root === null) {
return false;
}
// 如果到达叶子节点且当前路径和为目标和，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 $this->hasPathSum($root->left, $sum - $root->val) || $this->hasPathSum($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->right = new TreeNode(1);
$solution = new Solution();
$result = $solution->hasPathSum($root, 22);
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
class Solution:
def hasPathSum(self, root: TreeNode, sum: int) -> bool:
if not root:
return False
if not root.left and not root.right and root.val == sum:
return True
return self.hasPathSum(root.left, sum - root.val) or self.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.right = TreeNode(1)
solution = Solution()
result = solution.hasPathSum(root, 22)
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)
}
var Solution = function() {};
Solution.prototype.hasPathSum = function(root, sum) {
if (!root) {
return false;
}
if (!root.left && !root.right && root.val === sum) {
return true;
}
return this.hasPathSum(root.left, sum - root.val) || this.hasPathSum(root.right, sum - root.val);
};
// 示例用法
var 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.right = new TreeNode(1);
var solution = new Solution();
var result = solution.hasPathSum(root, 22);
console.log(result); // 输出 true
```
在以上代码中，我们定义了二叉树的节点结构，并实现了 `hasPathSum` 函数来检查是否存在路径和为目标和的情况。通过递归地检查每个节点，我们可以找到满足条件的路径。