寻找二叉树公共祖先结点递归实现

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
/*寻找公共祖先结点LCA算法de思想：
第零种情况：root为null时直接return
第一种情况：如果p在左子树q在右子树或反之则root为LCA
第二种情况：p和q均在左子树(LCA递归左子树)
第三种情况：p和q均在右子树(LCA递归右子树)
PS:最后return的right和left不会同时为NULL
*/
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root==NULL||root==p||root==q) return root;
        TreeNode* left=lowestCommonAncestor(root->left,p,q);
        TreeNode* right=lowestCommonAncestor(root->right,p,q);
        if(left && right){
            return root; 
        }else{
            return left==NULL?right:left;
        }
    }
};

/**
 * 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:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root == NULL || root == p || root == q){
            return root;
        }
        //将题目简化为查找以root为根结点的二叉树上是否有p或q节点,有则返回root
        //递归左子树
        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        //递归右子树
        TreeNode* right = lowestCommonAncestor(root->right, p, q);
        if(left && right){
            //左右子树上均能找到p或q，即p和q分居左右子树，直接返回root
            return root;
        }else{
            //最后的right和left不会同时为NULL
            return left == NULL? right: left;
        }
    }
};