236. Lowest Common Ancestor of a Binary Tree

 1 class Solution {
 2 public:
 3     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
 4         if (!root || root == p || root == q) return root; // 注意边界条件这里的意义
 5         else {
 6             TreeNode* left = lowestCommonAncestor(root->left, p, q);
 7             TreeNode* right = lowestCommonAncestor(root->right, p, q);
 8             if (left && right) return root; // 注意这里，左右都不为空，说明p，q分布在左右两个子树
 9             else if (! left) return right;
10             else return left;
11         }
12     }
13 };

 1 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
 2 // 深度优先遍历，生成两个单链表，求两个单链表的最后一个公共节点
 3 if (p == nullptr || q == nullptr) return (p == nullptr) ? q : p;
 4     vector<TreeNode*> vec1, vec2;
 5     TreeNode* curr = root;
 6     bool flag = false;
 7     while (true) { // 深搜构造链表
 8         while (curr != nullptr) {
 9             vec1.push_back(curr);
10             if (curr == p) {
11                 flag = true; break;
12             }
13             curr = curr->left;
14         }
15         if (curr != p) {
16             TreeNode* temp = new TreeNode(-1);
17             while (vec1.back()->right == nullptr || vec1.back()->right == temp) { // 后一个条件防止当前节点的右儿子是刚弹出的节点
18                 temp = vec1.back();
19                 vec1.pop_back();
20             }
21             delete temp;
22             curr = vec1.back()->right;
23         }
24         if (flag) break;
25     }
26     curr = root;
27     flag = false;
28     while (true) {
29         while (curr != nullptr) {
30             vec2.push_back(curr);
31             if (curr == q) {
32                 flag = true; break;
33             }
34             curr = curr->left;
35         }
36         if (curr != q) {
37             TreeNode* temp = new TreeNode(-1);
38             while (vec2.back()->right == nullptr || vec2.back()->right == temp) {
39                 temp = vec2.back();
40                 vec2.pop_back();
41             }
42             delete temp;
43             curr = vec2.back()->right;
44         }
45         if (flag) break;
46     }
47     int i = 0;
48     while (i < vec1.size() && i < vec2.size()) {
49         if(vec1[i]->val == vec2[i]->val) ++ i;
50         else break;
51     }
52     return vec1[i - 1];    
53 }