题目
给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。
示例
输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]
输入: preorder = [-1], inorder = [-1]
输出: [-1]
代码（递归法）
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
* right(right) {}
* };
*/
class Solution {
public:
unordered_map<int, int> index;
TreeNode* myBuildeTree(const vector<int>& preorder,
const vector<int>& inorder, int preorder_left,
int preorder_right, int inorder_left,
int inorder_right) {
if (preorder_left > preorder_right)
return nullptr;
//前序遍历的第一个节点就是根节点
int preorder_root = preorder_left;
//利用哈希表把根节点从中序遍历中找出来
int inorder_root = index[preorder[preorder_root]];
//先建立根节点
TreeNode* root = new TreeNode(preorder[preorder_root]);
//得到左子树节点数目
int size_left_subtree = inorder_root - inorder_left;
//递归构建左子树并连接到根节点
root->left = myBuildeTree(preorder, inorder,
preorder_left + 1,preorder_left + size_left_subtree,
inorder_left, inorder_root - 1);
//递归构建右子树并连接到根节点
root->right = myBuildeTree(preorder, inorder,
preorder_left + size_left_subtree + 1,
preorder_right, inorder_root + 1, inorder_right);
return root;
}
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
int n = preorder.size();
//对中序遍历的结果进行哈希映射
for (int i = 0; i < n; i++)
index[inorder[i]] = i;
return myBuildeTree(preorder, inorder, 0, n - 1, 0, n - 1);
}
};