一、题目说明
题目236. Lowest Common Ancestor of a Binary Tree,在一个二叉树中找两个节点的最近公共祖先。难度是Medium!
二、我的解答
这个用二叉树的递归遍历,稍加改造即可:
class Solution{
public:
TreeNode* lowestCommonAncestor(TreeNode* root,TreeNode*p,TreeNode*q){
if(root == NULL) return root;
if(root == p || root==q) return root;
TreeNode* left,*right;
left = lowestCommonAncestor(root->left,p,q);
right = lowestCommonAncestor(root->right,p,q);
if(left !=NULL && right!=NULL){
return root;
}else if(left != NULL){
return left;
}else if(right != NULL){
return right;
}else{
return NULL;
}
}
};
性能如下:
Runtime: 16 ms, faster than 94.88% of C++ online submissions for Lowest Common Ancestor of a Binary Tree.
Memory Usage: 16.7 MB, less than 87.27% of C++ online submissions for Lowest Common Ancestor of a Binary Tree.
三、优化措施
其他方法,暂时想不起来。