前/中序遍历序列重建二叉树

Construct Binary Tree from Preorder and Inorder Traversal

题意：输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

解题思路：取出前序遍历序列第一个结点的值，即根结点的值

在中序遍历序列中找出根结点的位置，计算distance(in_first, in_left_last)为len

则pre_left_last = next(pre_first, len + 1)

前序遍历序列中对应的左子树的前序遍历序列为next(pre_first), pre_left_last

中序遍历序列中对应的左子树的中序遍历序列为in_first, in_left_last

前序遍历序列中对应的右子树的前序遍历序列为pre_left_last, pre_last

中序遍历序列中对应的右子树的中序遍历序列为next(in_left_last), in_last

然后递归处理

/**
* 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>& inorder, vector<int>& postorder) {
return buildTree(begin(inorder), end(inorder), begin(postorder), end(postorder));
}

template<typename MyIt>
TreeNode* buildTree(MyIt in_first, MyIt in_last, MyIt post_first, MyIt post_last) {
if (in_first == in_last)
return nullptr;
if (post_first == post_last)
return nullptr;

const int rootValue = *prev(post_last);
auto in_left_last = find(in_first, in_last, rootValue);
int len = distance(in_first, in_left_last);
auto post_left_last = next(post_first, len);

TreeNode *root = new TreeNode(rootValue);
root->left = buildTree(in_first, in_left_last, post_first, post_left_last);
root->right = buildTree(next(in_left_last), in_last, post_left_last, prev(post_last));

return root;
}

};