[Leetcode]105. Construct Binary Tree from Preorder and Inorder Traversal

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本文介绍了一种使用前序遍历和中序遍历构建二叉树的方法。通过递归方式定位根节点，并根据中序遍历确定左右子树，最终完成整棵树的构建。

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

Note:

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

/**
 * 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) {
        return buildTree(begin(preorder), end(preorder), begin(inorder), end(inorder));
    }
    
    TreeNode* buildTree(vector<int>::iterator pre_first, vector<int>::iterator pre_last, vector<int>::iterator in_first, vector<int>::iterator in_last) {
        if (pre_first == pre_last)
            return nullptr;
        if (in_first == in_last)
            return nullptr;
        auto root = new TreeNode(*pre_first);
        auto inrootPos = find(in_first, in_last, *pre_first);
        auto leftLen = distance(in_first, inrootPos);
        root->left = buildTree(next(pre_first), next(pre_first, leftLen + 1), in_first, next(in_first, leftLen));
        root->right = buildTree(next(pre_first, leftLen + 1), pre_last, next(inrootPos), in_last);
        return root;
    }
};