# 【LeetCode】105. Construct Binary Tree from Preorder and Inorder Traversal

有參考網路上各位神人的寫法，不過我是靠自己把解題思路想出來，對於left and right的上下界線不是那麼清楚。

/**
 * 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>& preorder, vector<int>& inorder) {
        return build(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
    }
    TreeNode* build(vector<int>& pre, int pL, int pR, vector<int>& in, int iL, int iR){
        if (iL > iR || pL > pR){
            return NULL;
        }
        int mid = 0;
        for (mid = iL; mid <= iR; ++mid) {
            if (pre[pL] == in[mid]) break;
        }
        TreeNode *root = new TreeNode(pre[pL]);
        root->left = build(pre, pL+1, pL+mid-iL, in, iL, mid-1);
        root->right = build(pre, pL-iL+mid+1, pR, in, mid+1, iR);
        
        return root;
        
    }
};

###### tags: `LeetCode` `程式`