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| 1 | +想法:因為是一個preorder 的序列,所以我們可以用一個 array 紀錄每一層上一個插入的節點 |
| 2 | +接著,在字串中把每個數字提取出來,同時紀錄中間相隔了多少 '-' (代表這個數字要被插入的深度) |
| 3 | +只需將 array 中上一層的節點下新增子節點即可 |
| 4 | + |
| 5 | +Time Complexity : O(n) for traversing the string |
| 6 | +Space Complexity : O(h) = O(n) for the array to store the last insert node of each height |
| 7 | + |
| 8 | +/** |
| 9 | + * Definition for a binary tree node. |
| 10 | + * struct TreeNode { |
| 11 | + * int val; |
| 12 | + * TreeNode *left; |
| 13 | + * TreeNode *right; |
| 14 | + * TreeNode() : val(0), left(nullptr), right(nullptr) {} |
| 15 | + * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} |
| 16 | + * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} |
| 17 | + * }; |
| 18 | + */ |
| 19 | +class Solution { |
| 20 | +public: |
| 21 | + TreeNode* recoverFromPreorder(string traversal) { |
| 22 | + vector<TreeNode*> lastinserted ; |
| 23 | + int i = 0 ; |
| 24 | + int rootnumber = traversal[i] - '0' ; |
| 25 | + while ( i + 1 < traversal.length() && traversal[i + 1] != '-' ) { |
| 26 | + rootnumber *= 10 ; |
| 27 | + rootnumber += traversal[i + 1] - '0' ; |
| 28 | + i++ ; |
| 29 | + } |
| 30 | + |
| 31 | + lastinserted.push_back( new TreeNode(rootnumber) ) ; |
| 32 | + |
| 33 | + int height = 0; |
| 34 | + for(i = i + 1; i < traversal.length() ; i++) { |
| 35 | + if (traversal[i] != '-') { |
| 36 | + //cout << height << " " << i << endl ; |
| 37 | + int number = traversal[i] - '0' ; |
| 38 | + while ( i + 1 < traversal.length() &&traversal[i + 1] != '-' ) { |
| 39 | + number *= 10 ; |
| 40 | + number += traversal[i + 1] - '0' ; |
| 41 | + i++; |
| 42 | + } |
| 43 | + |
| 44 | + |
| 45 | + TreeNode* nownode = new TreeNode(number); |
| 46 | + if ( lastinserted[height - 1]->left == nullptr ) { |
| 47 | + lastinserted[height - 1]->left = nownode ; |
| 48 | + } |
| 49 | + else |
| 50 | + lastinserted[height - 1]->right = nownode ; |
| 51 | + |
| 52 | + if ( height >= lastinserted.size() ) |
| 53 | + lastinserted.push_back( nownode ) ; |
| 54 | + else |
| 55 | + lastinserted[height] = nownode ; |
| 56 | + height = 0 ; |
| 57 | + } |
| 58 | + else { |
| 59 | + height++ ; |
| 60 | + } |
| 61 | + |
| 62 | + } |
| 63 | + return lastinserted[0]; |
| 64 | + } |
| 65 | +}; |
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