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| 1 | +/** |
| 2 | + * Definition for a binary tree node. |
| 3 | + * struct TreeNode { |
| 4 | + * int val; |
| 5 | + * TreeNode *left; |
| 6 | + * TreeNode *right; |
| 7 | + * TreeNode() : val(0), left(nullptr), right(nullptr) {} |
| 8 | + * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} |
| 9 | + * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} |
| 10 | + * }; |
| 11 | + */ |
| 12 | +class Solution { |
| 13 | +public: |
| 14 | + string getDirections(TreeNode* root, int startValue, int destValue) |
| 15 | + { |
| 16 | + vector<int>nums1, nums2; |
| 17 | + nums1.push_back(root->val); |
| 18 | + nums2.push_back(root->val); |
| 19 | + string dirs1, dirs2; |
| 20 | + dfs(root, nums1, dirs1, startValue); |
| 21 | + dfs(root, nums2, dirs2, destValue); |
| 22 | + |
| 23 | + int k = 0; |
| 24 | + while (k<nums1.size() && k<nums2.size() && nums1[k]==nums2[k]) |
| 25 | + k++; |
| 26 | + k--; |
| 27 | + for (int i=k; i<dirs1.size(); i++) dirs1[i] = 'U'; |
| 28 | + return dirs1.substr(k)+dirs2.substr(k); |
| 29 | + |
| 30 | + } |
| 31 | + |
| 32 | + bool dfs(TreeNode* node, vector<int>&nums, string&dirs, int target) |
| 33 | + { |
| 34 | + if (node==NULL) return false; |
| 35 | + if (node->val == target) return true; |
| 36 | + if (node->left) |
| 37 | + { |
| 38 | + nums.push_back(node->left->val); |
| 39 | + dirs.push_back('L'); |
| 40 | + if (dfs(node->left, nums, dirs, target)) |
| 41 | + return true; |
| 42 | + nums.pop_back(); |
| 43 | + dirs.pop_back(); |
| 44 | + } |
| 45 | + if (node->right) |
| 46 | + { |
| 47 | + nums.push_back(node->right->val); |
| 48 | + dirs.push_back('R'); |
| 49 | + if (dfs(node->right, nums, dirs, target)) |
| 50 | + return true; |
| 51 | + nums.pop_back(); |
| 52 | + dirs.pop_back(); |
| 53 | + } |
| 54 | + return false; |
| 55 | + } |
| 56 | + |
| 57 | +}; |
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