题目：236. 二叉树的最近公共祖先

思路

代码

用深度搜索的思想（好吧，前序、中序、后序都是深搜思想），保存寻找路径，看看找到2个节点的路径的重合部分，就可以找到最近公共祖先；

/*** 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:vector<TreeNode*> path;vector<vector<TreeNode*>> result;void travel(TreeNode* cur, TreeNode* target){   if(cur == NULL){return;}// 中if(cur->val == target->val){result.push_back(path);return;}// 左if(cur->left){path.push_back(cur->left);travel(cur->left, target);path.pop_back();}// 右if(cur->right){path.push_back(cur->right);travel(cur->right, target);path.pop_back();}}TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {int i, j, min_length;path.push_back(root);travel(root, q);path.clear();path.push_back(root);travel(root, p);/*for(i = 0; i < result.size(); i++){for(j = 0; j < result[i].size(); j++){cout << result[i][j]->val << ", ";}cout << endl;}*/// cout << "result[0].size() : " << result[0].size() << endl;// cout << "result[1].size() : " << result[1].size() << endl;min_length = min(result[0].size(), result[1].size());for(i = 0; i < min_length; i++){if(result[0][i]->val == result[1][i]->val){continue;}else{break;}}// cout << "i : " << i << "result[0][i] : " << result[0][i]->val;return result[0][i-1];}
};