Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to thedefinition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes pand qas the lowest node in T that has both pand qas descendants (where we allowa 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
• AllNode.valareunique.
• p != q
• pandqwillexist in the tree.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 
 
*/
class Solution {
public:
    //還是遞迴解法：在左子樹中遞迴找p或q，如果找到一個就返回；在右子樹中找p或q如果找到一個就返回那個節點，沒找到就返回NULL。
    //如果左子樹中找到一個，右子樹中也找到一個，那麼這兩個子樹的根節點就是結果。
    //如果左子樹中沒有找到，那麼返回空，說明兩個節點都在右子樹。否則都在左子樹。
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root==NULL) return NULL;
        if(root==p||root==q)  return root;
        TreeNode* ltree=lowestCommonAncestor(root->left,p,q);
        TreeNode* rtree=lowestCommonAncestor(root->right,p,q);
        //分別在左右子樹中找到
        if(ltree&&rtree)  return root;
        //右三種情況在此處返回
        //在左右某個子樹中沒找到任一個p或q
        //第一次在當前節點左子樹或者右子樹中找到p或q
        //當前節點的左右子樹找到p'q
        return  ltree?ltree:rtree;
    }
};