leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal

构建二叉树

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本文介绍了一种通过前序遍历和中序遍历构建二叉树的方法。利用递归方式寻找根节点，并根据根节点划分左右子树，最终构建完整的二叉树结构。

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

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

在中序遍历中找前序遍历的第一个，也就是根节点。

找到的这个值将中序遍历分为左子树和右子树。

/**
 * 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* helper(vector<int> &preorder, int begin1, int end1, vector<int> &inorder, int begin2, int end2)
    {
        if (begin1 > end1)
            return NULL; 
        int lenleft = find(inorder.begin(), inorder.end(), preorder[begin1]) - inorder.begin() - begin2;
        
        TreeNode *p = new TreeNode(preorder[begin1]);     
        p->left = helper(preorder, begin1 + 1, begin1 + lenleft, inorder, begin2, begin2 + lenleft - 1);
        p->right = helper(preorder, begin1 + lenleft + 1, end1, inorder, begin2 + lenleft + 1, end2);
        return p;
    }
    
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) 
    {
        return helper(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);
    }
};