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

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

有了先序和中序，基本思路就是遞迴的找左右子樹的根節點，加入到樹中
/**
 * 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 *root = NULL;  // 儲存樹的根節點
    // 遞迴函式，pre是先序，pl和pr為左或右子樹的區間，in為中序，il和ir同為左或右子樹的區間，sub決定是左子樹還是右子樹，par為父親結點
    void fun(vector<int> &pre, int pl, int pr, vector<int> &in, int il, int ir, int sub, TreeNode *par) {
        if (pl <= pr) { // 左或右子樹不為空
 

            TreeNode *tmp = new TreeNode(pre[pl]); // 新建一個結點，即某個子樹的根節點
            if (root == NULL) {
                root = tmp; // 根節點
                par = root;
            } else {
                // 根據sub判斷新增到左子樹還是右子樹
                if (sub == 0) par->left = tmp;
                if (sub == 1) par->right = tmp;
            }

            // 找到根節點在中序遍歷中的位置
 

            int ipos = il;
            for (int i = il; i <= ir; ++i) {
                if (in[i] == pre[pl]) {
                    ipos = i;
                    break;
                }
            }

            // 找到先序遍歷中左右子樹的分隔位置
            int ppos = pl + ipos - il;

            // 遞迴構造左右子樹
            if (ipos > il) fun(pre, pl+1, ppos, in, il, ipos-1, 0, tmp); // 有左子樹
            if (ipos < ir) fun(pre, ppos+1, pr, in, ipos+1, ir, 1, tmp); // 有右子樹
        }
    }    

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if (preorder.size() == 0) return NULL;
        //TreeNode *root = NULL;
        fun(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1, 0, NULL);
        return root;

    }
};