Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
这个题目的解法与LeetCode[Tree]: Construct Binary Tree from Preorder and Inorder Traversal的解法几乎相差无几。我的C++代码实现如下:
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; } };
时间性能如下图所示:
时间: 2024-10-09 02:55:29