用两个变量,回溯时判断即可。
/**
* 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* ans;
void dfs(TreeNode* now,TreeNode* p,TreeNode* q,bool& flagP,bool& flagQ){
bool A = false;
bool B = false;
if(now->left != NULL) dfs(now->left,p,q,A,B);
if(now->right != NULL) dfs(now->right,p,q,A,B);
if(now == p) A = true;
if(now == q) B = true;
if(ans == NULL && A && B) ans = now;
flagP = flagP || A;
flagQ = flagQ || B;
}
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
bool flagP = false;
bool flagQ = false;
ans = NULL;
dfs(root,p,q,flagP,flagQ);
if(ans == NULL && flagP == flagQ) ans = root;
return ans;
}
};