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中序遍历的根节点左侧是左子树，右侧是右子树，后序遍历的最后一个元素为根节点。

在中序遍历中找到根节点，从而找到左右子树，知道左右子树的范围，从而后序遍历中的左右子树也就确定好了。

然后分别对左右子树用同样的方式递归构造下去。

/*** 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:TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {unordered_map<int, int> inorder_map;for(int i = 0; i < inorder.size(); i++){inorder_map[inorder[i]] = i;}return buildTreeHelper(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1, inorder_map);}
private:TreeNode* buildTreeHelper(vector<int>& inorder, int in_start, int in_end, vector<int>& postorder, int post_start, int post_end, unordered_map<int, int>& inorder_map){if(in_start > in_end || post_start > post_end) return nullptr;int rootVal = postorder[post_end];TreeNode* root = new TreeNode(rootVal);int midIndex = inorder_map[rootVal];int leftTreeSize = midIndex - in_start;root->left = buildTreeHelper(inorder, in_start, midIndex - 1, postorder, post_start, post_start + leftTreeSize - 1, inorder_map);root->right = buildTreeHelper(inorder, midIndex + 1, in_end, postorder, post_start + leftTreeSize, post_end - 1, inorder_map);return root;}
};