当前位置： 面试刷题>> 寻找树中最左下节点的值 (经典算法题500道)
### 题目描述补充
给定一棵二叉树，请编写一个函数来找到并返回树中最左下节点的值。最左下节点是指在所有可能的左子树路径中，深度最深的左节点。如果树为空，则返回某个特定的值（例如，`null`、`None`、`undefined` 或特定的错误码等，具体取决于编程语言的习惯）。
### 示例
假设有以下二叉树：
```
1
/ \
2 3
/ \
4 5
/
6
```
最左下节点的值为 6。
### PHP 示例代码
```php
val = $val;
$this->left = $left;
$this->right = $right;
}
}
function findLeftmostValue($root) {
if ($root === null) {
return null; // 或者返回某个特定的错误码
}
$result = $root->val;
$current = $root;
while ($current->left !== null) {
$result = $current->left->val;
$current = $current->left;
}
return $result;
}
// 示例用法
$root = new TreeNode(1);
$root->left = new TreeNode(2);
$root->right = new TreeNode(3);
$root->left->left = new TreeNode(4);
$root->left->right = new TreeNode(5);
$root->left->right->left = new TreeNode(6);
echo findLeftmostValue($root); // 输出 6
?>
```
### 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 findLeftmostValue(root):
if not root:
return None # 或者返回某个特定的错误码
result = root.val
current = root
while current.left:
result = current.left.val
current = current.left
return result
# 示例用法
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
root.left.right.left = TreeNode(6)
print(findLeftmostValue(root)) # 输出 6
```
### 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 findLeftmostValue(root) {
if (!root) {
return null; // 或者返回某个特定的错误码
}
let result = root.val;
let current = root;
while (current.left) {
result = current.left.val;
current = current.left;
}
return result;
}
// 示例用法
let root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(5);
root.left.right.left = new TreeNode(6);
console.log(findLeftmostValue(root)); // 输出 6
```
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