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# LeetCode236 lowest Common Ancestor of a binary tree（二叉树的最近公共祖先）

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### 题目

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).

`````` Example 1:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant
of itself according to the LCA definition.

Example 3:
Input: root = [1,2], p = 1, q = 2
Output: 1

Constraints:
The number of nodes in the tree is in the range [2, 10⁵].
-10⁹ <= Node.val <= 10⁹
All Node.val are unique.
p != q
p and q will exist in the tree.
``````

### 方法

#### 深度优先遍历法（DFS）

• 时间复杂度：O(n)
• 空间复杂度：O(n)
``````class Solution {
private TreeNode result = null;
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
dfs(root,p,q);
return result;
}
private boolean dfs(TreeNode root, TreeNode p, TreeNode q){
if(root==null){
return false;
}
boolean left = dfs(root.left,p,q);
boolean right = dfs(root.right,p,q);
if(left&&right || ((left||right)&&(root.val==p.val||root.val==q.val))) {
result = root;
}
return left || right || (root.val==p.val || root.val==q.val);
}
}
``````

• 时间复杂度：O(n)
• 空间复杂度：O(n)