Lowest Common Ancestor of a Binary Tree

1，题目要求

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.

给定二叉树，找到树中两个给定节点的最低共同祖先（LCA）。

根据维基百科上LCA的定义：“最低共同祖先在两个节点p和q之间定义为T中的最低节点，其中p和q都是后代（我们允许节点成为其自身的后代）。”

2，题目思路

对于这道题，是寻找两个节点在树中的最低共同祖先。

在解决上，我们可以使用递归的策略，分别在子树中寻找节点p和节点q。

对于一个节点root：

如果它的左右子树中都没有p和q，就说明它不是二者的祖先。
如果p和q分别在root的左右子树/右左子树中，说明root就是p和q的祖先，而且应该也是最低公共祖先。
如果在左子树中没有搜到p或者q，但是右子树搜到了，就说明root->right是p或者q的祖先。右子树同理。

3，代码实现

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/int x = []() {ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);return 0;
}();class Solution {public:TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if(root == nullptr || root == p || root == q)return root;TreeNode* left = lowestCommonAncestor(root->left, p ,q);TreeNode* right = lowestCommonAncestor(root->right, p, q);if(left == nullptr && right == nullptr)return nullptr;if(left != nullptr && right != nullptr)return root;return left == nullptr? right : left;}
};