[106]从中序与后序遍历序列构造二叉树

根据一棵树的中序遍历与后序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。
例如，给出

中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]

返回如下的二叉树：

3
   / \
  9  20
    /  \
   15   7

/**
 * Definition for a binary tree node.
 * 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) {
        return build(inorder, 0, inorder.size() - 1,
                     postorder, 0, postorder.size() - 1);
    }

    TreeNode* build(vector<int>& inorder,  int i_start, int i_end,
                    vector<int>& postorder, int p_start, int p_end)
    {
        if(p_start > p_end){
            return NULL;
        }
        int root_val = postorder[p_end];
        int index = i_start;
        for(int i = i_start; i<inorder.size();i++){
            if(root_val == inorder[i]){
                index = i;
                break;
            }
        }
        int left_size = index - i_start;
        TreeNode* root = new TreeNode(root_val);
        root->left = build(inorder, i_start, index - 1,
                           postorder, p_start, p_start + left_size - 1);
        root->right = build(inorder, index + 1, i_end,
                            postorder, p_start + left_size, p_end - 1);
        return root;
    }
};