leetcode 236. 二叉树的最近公共祖先 dfs

从上之下一次找p和q的祖先链条，一直找到root
分别存在数组里面
然后找公共的祖先的最小的那个

#define debug(x) cout<<#x<<": "<<(x)<<endl;

class Solution {
public:
    vector<TreeNode*> pf;
    vector<TreeNode*> qf;
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        rootIsFatherOf(qf,root,q);
        rootIsFatherOf(pf,root,p);
        if(pf.size() == 0 || qf.size() == 0){
            return root;
        }
        int i = pf.size()-1;
        int j = qf.size()-1;
        TreeNode * ret = nullptr;
        for(;i>=0 && j>=0;){
            if(pf[i] == qf[j]){
                ret = pf[i];
                i--,j--;
            }else{
                break;
            }
        }
        if(ret == nullptr){
            return root;
        }
        return ret;
    }
    
    bool rootIsFatherOf(vector<TreeNode*> &ve,TreeNode* root,TreeNode* p){
        if(root == nullptr){
            return false;
        }
        if(root == p){
            return true;
        }
        else if( rootIsFatherOf(ve,root->left,p) ){
            ve.push_back(root->left);
            return true;
        }
        else if( rootIsFatherOf(ve,root->right,p) ){
            ve.push_back(root->right);
            return true;
        }else{
            return false;
        }
        return false;
    }
};

递归

答案参考: ​​https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree/comments/​​

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;
        else if(left != nullptr && right != nullptr) return root;
        else return left == nullptr ? right : left;
    }
    
};