[LeetCode] C++: Intermediate Problem-Tree 106. Construct a binary tree from the middle and post-order traversal sequence

106. Construct a binary tree from middle order and post order traversal sequence

Medium difficulty 462

The binary tree is constructed according to the middle-order traversal and post-order traversal of a tree.

Note:
You can assume that there are no duplicate elements in the tree.

For example, given

Inorder traversal inorder = [9,3,15,20,7]
Postorder traversal postorder = [9,15,7,20,3]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

Similar to the pre-order traversal, the post-order traversal is over, from back to front

/**
 * 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:
    int post_size;
    unordered_map<int, int> index;
    TreeNode* myBuildTree(const vector<int>& inorder, const vector<int>& postorder, int in_left, int in_right){
        if(in_left > in_right){
            return nullptr;
        }

        int root_val = postorder[post_size];
        TreeNode* root = new TreeNode(root_val);

        int id = index[root_val];
        post_size--;
        root->right = myBuildTree(inorder, postorder, id+1, in_right);
        root->left = myBuildTree(inorder, postorder, in_left, id-1);

        return root;
    }

    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        post_size = postorder.size()-1;
        for(int i = 0; i < inorder.size(); i++){
            index[inorder[i]] = i;
        }
        return myBuildTree(inorder, postorder, 0, inorder.size()-1);
    }
};