106. Construct Binary Tree from Inorder and Postorder Traversal (Tree; DFS)

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        root = NULL;
        if(inorder.empty()) return root;
        root = new TreeNode(0);
        buildSubTree(inorder,postorder,0,inorder.size()-1,0,postorder.size()-1,root);
        return root;

    }
    void buildSubTree(vector<int> &inorder, vector<int>&postorder, int inStartPos, int inEndPos, int postStartPos, int postEndPos,TreeNode * currentNode)
    {
        currentNode->val = postorder[postEndPos]; //后序遍历的最后一个节点是根节点

        //find root position in inorder vector
        int inRootPos;
        for(int i = inStartPos; i <= inEndPos; i++)
        {
            if(inorder[i] == postorder[postEndPos])
            {
                inRootPos = i;
                break;
            }
        }

        //right tree: 是中序遍历根节点之后的部分，对应后序遍历根节点前相同长度的部分
        int newPostPos = postEndPos - max(inEndPos - inRootPos, 0);
        if(inRootPos<inEndPos)
        {
            currentNode->right = new TreeNode(0);
            buildSubTree(inorder,postorder,inRootPos+1,inEndPos,newPostPos,postEndPos-1,currentNode->right);
        }

        //leftTree: 是中序遍历根节点之前的部分，对应后序遍历从头开始相同长度的部分
        if(inRootPos>inStartPos)
        {
            currentNode->left = new TreeNode(0);
            buildSubTree(inorder,postorder,inStartPos,inRootPos-1,postStartPos,newPostPos-1,currentNode->left);
        }
    }
private:
    TreeNode* root;
};