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).”

Given the following binary tree:  root = [3,5,1,6,2,0,8,null,null,7,4]

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.

Note:

• All of the nodes' values will be unique.
• p and q are different and both values will exist in the binary tree.

給定一個二叉樹, 找到該樹中兩個指定節點的最近公共祖先。

百度百科中最近公共祖先的定義為：“對於有根樹 T 的兩個結點 p、q，最近公共祖先表示為一個結點 x，滿足 x 是 p、q 的祖先且 x 的深度儘可能大（一個節點也可以是它自己的祖先）。”

例如，給定如下二叉樹:  root = [3,5,1,6,2,0,8,null,null,7,4]

示例 1:

輸入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
輸出: 3
解釋: 節點 5 和節點 1 的最近公共祖先是節點 3。

示例 2:

輸入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
輸出: 5
解釋: 節點 5 和節點 4 的最近公共祖先是節點 5。因為根據定義最近公共祖先節點可以為節點本身。

說明:

• 所有節點的值都是唯一的。
• p、q 為不同節點且均存在於給定的二叉樹中。

遞迴

 1 class Solution {
 2     func lowestCommonAncestor(_ root: TreeNode?, _ p: TreeNode?,_ q: TreeNode?) -> TreeNode? {
 3         if root!.val == nil || root!.equal(p) || root!.equal(q)
 4         {
 5             return root
 6         }
 7         
 8         var left:TreeNode? = lowestCommonAncestor(root!.left, p, q)
 9         if left!.val != nil && left!.equal(p) && left!.equal(q)
10         {
11             return left
12         }
13         
14         var right:TreeNode? = lowestCommonAncestor(root!.right, p , q)
15         if left!.val != nil && right!.val != nil
16         {
17             return root
18         }
19         return left!.val != nil ? left : right
20     }
21 }
22 public class TreeNode {
23     public var val: Int
24      public var left: TreeNode?
25      public var right: TreeNode?
26      public init(_ val: Int) {
27         self.val = val
28         self.left = nil
29         self.right = nil
30     }
31     
32     func equal(_ root: TreeNode?)-> Bool 
33     {
34         return (self.val == root!.val) && (self.left!.val == root!.left!.val) && (self.right!.val == root!.right!.val)
35     }    
36  }

C++：4ms

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 static const auto _____ = []()
11 {
12     ios::sync_with_stdio(false);
13     cin.tie(nullptr);
14     return nullptr;
15 }();
16 class Solution {
17 public:
18     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
19         if (root == NULL || root == p || root == q) return root;
20         TreeNode* ptr1 = lowestCommonAncestor(root->left, p, q);
21         TreeNode* ptr2 = lowestCommonAncestor(root->right, p, q);
22         if (ptr1 && ptr2) return root;
23         return ptr1 ? ptr1 : ptr2;
24     }
25 };

C++：12ms

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
13         if(!root||p==root||q==root) return root;
14         TreeNode *left = lowestCommonAncestor(root->left, p, q);
15         TreeNode *right = lowestCommonAncestor(root->right, p, q);
16         if (left && right) return root;
17         return left ? left : right;
18         
19     }
20 };