输入某二叉树的前序遍历和中序遍历的结果,请构建该二叉树并返回其根节点。
假设输入的前序遍历和中序遍历的结果中都不含重复的数字。
示例 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7] Output: [3,9,20,null,null,15,7]
示例 2:
Input: preorder = [-1], inorder = [-1] Output: [-1]
限制:
0 <= 节点个数 <= 5000
- #include
- #include
- #include
-
- using namespace std;
-
- struct TreeNode
- {
- int val;
- TreeNode *left;
- TreeNode *right;
- TreeNode(int x) : val(x), left(NULL), right(NULL) {}
- };
-
- class Solution {
- public:
- TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
- for (int i = 0; i < inorder.size(); i++)
- {
- // 用中序遍历数组建立 值-下标 的映射
- value_index[inorder[i]] = i;
- }
- return recursive(preorder,0, 0, inorder.size() - 1);
- }
- TreeNode *recursive(vector<int>& pre,int pre_root, int in_left, int in_right)
- {
- if (in_left > in_right) return nullptr;
- // 将对应的 val 赋给 node 节点
- TreeNode *node = new TreeNode(pre[pre_root]);
- int in_root = value_index[pre[pre_root]];
- node->left = recursive(pre,pre_root + 1, in_left, in_root - 1);
- node->right = recursive(pre,pre_root + in_root - in_left + 1, in_root + 1, in_right);
- return node;
- }
- private:
- unordered_map<int, int> value_index;
- };