(C++)LeetCode#105. Construct Binary Tree from Preorder and Inorder Traversal

• 题目：根据二叉树的前序和后序遍历结果，重新构造二叉树
• 难度：Medium
• 思路：遍历前序结果，找到根节点在中序遍历中的index，然后继续构造左右子树。【确定左右子树的元素所在区间，包括前序区间和中序区间】
• 代码：

/**
 * 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>& preorder, vector<int>& inorder) {
        map<int, int> m;
        if(preorder.empty()){
            return NULL;
        }
        int inlen = inorder.size();
        int prelen = preorder.size();
        if(inlen == 0 || prelen == 0 || inlen != prelen){
            return NULL;
        }

        return helper(preorder, 0, inlen-1, inorder, 0, inlen-1);
    }

    TreeNode* helper(vector<int>& preorder, int ps, int pe, vector<int>& inorder, int is, int ie){
        if(ps > pe || is > ie){
            return NULL;
        }
        TreeNode* root = new TreeNode(preorder[ps]);
        auto f = find(inorder.begin() + is, inorder.begin() + ie, preorder[ps]);
        //int ir = f - inorder.begin(); 此方式也ok
        int ir = distance(inorder.begin(), f);
        root->left = helper(preorder, ps+1, ps+ir-is, inorder, is, ir-1);
        root->right = helper(preorder, ps+ir-is+1, pe, inorder, ir+1, ie);
        return root;
    }
};