2026/03/27

[LeetCode] 572 Subtree of Another Tree

當val一樣時

如果已經固定了, 就一定要完全一樣到底

如果尚未固定, 則有兩種選擇

用當前root, 之後固定

不用當前root, 維持不固定

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  bool check(TreeNode* root, TreeNode* subRoot, bool fixed) {
    if (root == nullptr && subRoot == nullptr)
      return fixed;
    if (root == nullptr || subRoot == nullptr)
      return false;
    if (root->val != subRoot->val) 
      if (fixed)
        return false;
      else 
        return check(root->left, subRoot, false) ||
               check(root->right, subRoot, false);
    if (fixed)
      return check(root->left, subRoot->left, fixed) && 
             check(root->right, subRoot->right, fixed);

    bool tryRoot = check(root->left, subRoot->left, true) && 
                   check(root->right, subRoot->right, true);
    if (tryRoot)
      return true;
    
    return check(root->left, subRoot, false) ||
           check(root->right, subRoot, false);
  }

  bool isSubtree(TreeNode* root, TreeNode* subRoot) {
    return check(root, subRoot, false);
  }
};
Labels: