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

• Title: Reconstruct the binary tree according to the results of the preorder and postorder traversal of the binary tree
• Difficulty: Medium
• Idea: Traverse the pre-order results, find the index of the root node in the in-order traversal, and then continue to construct the left and right subtrees. [Determine the interval where the elements of the left and right subtrees are located, including the preorder interval and the middle order interval]
• Code:

/**
 * 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;
    }
};