Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
這個題目的解法與LeetCode[Tree]: Construct Binary Tree from Preorder and Inorder Traversal的解法幾乎相差無幾。我的C++代碼實作如下:
class Solution {
public:
TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
return buildTreeHelper(inorder.begin(), inorder.end(), postorder.begin(), postorder.end());
}
TreeNode *buildTreeHelper(vector<int> ::iterator inBegin, vector<int> ::iterator inEnd, vector<int> ::iterator postBegin, vector<int> ::iterator postEnd) {
if (postEnd <= postBegin) return nullptr;
TreeNode *root = new TreeNode(*(postEnd - 1));
int leftNodes = find(inBegin, inEnd, *(postEnd - 1)) - inBegin;
root->left = buildTreeHelper(inBegin, inBegin + leftNodes, postBegin, postBegin + leftNodes);
root->right = buildTreeHelper(inBegin + leftNodes + 1, inEnd, postBegin + leftNodes, postEnd - 1);
return root;
}
};
時間性能如下圖所示:
![](https://img.laitimes.com/img/_0nNw4CM6IyYiwiM6ICdiwiIyVGduV2QvwVe0lmdhJ3ZvwFM38CXlZHbvN3cpR2Lc1TPB10QGtWUCpEMJ9CXsxWam9CXwADNvwVZ6l2c052bm9CXUJDT1wkNhVzLcRnbvZ2Lc1TP350dBRlWthmMZZXUYpVd1kmYr50MZV3YyI2cKJDT29GRjBjUIF2LcRHelR3LcJzLctmch1mclRXY39zN4cDN0QTMwADOyITM0EDMy8CX0Vmbu4GZzNmLn9Gbi1yZtl2Lc9CX6MHc0RHaiojIsJye.jpg)