[leetcode 105] Construct Binary Tree from Preorder and Inorder Traversal

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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。利用递归技术定位根节点，并根据中序遍历划分左右子树，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 binary tree
 * 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 build(begin(preorder), end(preorder), begin(inorder), end(inorder));
    }
    template<typename Inter>
    TreeNode *build(Inter pre_first, Inter pre_end, Inter in_first, Inter in_end) {
        if (pre_first == pre_end || in_first == in_end) {
            return NULL;
        }
        TreeNode *root = new TreeNode(*pre_first);
        auto inRootPos = find(in_first, in_end, *pre_first);
        auto leftSize = distance(in_first, inRootPos);
        root->left = build(next(pre_first), next(pre_first, leftSize+1), in_first, next(in_first, leftSize));
        root->right = build(next(pre_first, leftSize+1), pre_end, next(in_first, leftSize+1), in_end);
        return root;
    }
};