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feat: add python and java solutions to lcci question
添加《程序员面试金典》题解：面试题 04.08. 首个共同祖先
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‎lcci/04.08.First Common Ancestor/README.md

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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:
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if root is None or root == p or root == q:
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return root
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left = self.lowestCommonAncestor(root.left, p, q)
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right = self.lowestCommonAncestor(root.right, p, q)
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return right if left is None else (left if right is None else root)
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```
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### Java
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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if (root == null || root == p || root == q) {
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return root;
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}
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TreeNode left = lowestCommonAncestor(root.left, p, q);
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TreeNode right = lowestCommonAncestor(root.right, p, q);
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return left == null ? right : (right == null ? left : root);
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}
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}
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```
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### ...

‎lcci/04.08.First Common Ancestor/README_EN.md

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# [04.08. First Common Ancestor](https://leetcode-cn.com/problems/first-common-ancestor-lcci)
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## Description
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<p>Design an algorithm and write code to find the first common ancestor of two nodes in a binary tree. Avoid storing additional nodes in a data structure. NOTE: This is not necessarily a binary search tree.</p>
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<p>For example, Given the following tree: root = [3,5,1,6,2,0,8,null,null,7,4]</p>
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<pre>
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3
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/ \
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5 1
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/ \ / \
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6 2 0 8
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/ \
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7 4
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</pre>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
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<strong>Input:</strong> 3
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<strong>Explanation:</strong> The first common ancestor of node 5 and node 1 is node 3.</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
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<strong>Output:</strong> 5
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<strong>Explanation:</strong> The first common ancestor of node 5 and node 4 is node 5.</pre>
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<p><strong>Notes:</strong></p>
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<ul>
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<li>All node values are pairwise distinct.</li>
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<li>p, q are different node and both can be found in the given tree.</li>
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</ul>
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## Solutions
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### Python3
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```python
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```
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### Java
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```java
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```
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### ...
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```
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```
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# [04.08. First Common Ancestor](https://leetcode-cn.com/problems/first-common-ancestor-lcci)
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## Description
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<p>Design an algorithm and write code to find the first common ancestor of two nodes in a binary tree. Avoid storing additional nodes in a data structure. NOTE: This is not necessarily a binary search tree.</p>
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<p>For example, Given the following tree: root = [3,5,1,6,2,0,8,null,null,7,4]</p>
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<pre>
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3
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/ \
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5 1
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/ \ / \
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6 2 0 8
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/ \
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7 4
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</pre>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
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<strong>Input:</strong> 3
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<strong>Explanation:</strong> The first common ancestor of node 5 and node 1 is node 3.</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
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<strong>Output:</strong> 5
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<strong>Explanation:</strong> The first common ancestor of node 5 and node 4 is node 5.</pre>
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<p><strong>Notes:</strong></p>
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<ul>
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<li>All node values are pairwise distinct.</li>
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<li>p, q are different node and both can be found in the given tree.</li>
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</ul>
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## Solutions
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### Python3
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:
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if root is None or root == p or root == q:
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return root
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left = self.lowestCommonAncestor(root.left, p, q)
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right = self.lowestCommonAncestor(root.right, p, q)
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return right if left is None else (left if right is None else root)
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```
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### Java
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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if (root == null || root == p || root == q) {
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return root;
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}
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TreeNode left = lowestCommonAncestor(root.left, p, q);
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TreeNode right = lowestCommonAncestor(root.right, p, q);
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return left == null ? right : (right == null ? left : root);
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}
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}
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```
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### ...
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```
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```

‎lcci/04.08.First Common Ancestor/Solution.java

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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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if (root == null || root == p || root == q) {
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return root;
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}
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TreeNode left = lowestCommonAncestor(root.left, p, q);
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TreeNode right = lowestCommonAncestor(root.right, p, q);
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return left == null ? right : (right == null ? left : root);
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}
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}

‎lcci/04.08.First Common Ancestor/Solution.py

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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:
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if root is None or root == p or root == q:
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return root
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left = self.lowestCommonAncestor(root.left, p, q)
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right = self.lowestCommonAncestor(root.right, p, q)
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return right if left is None else (left if right is None else root)

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