[Leetcode] 106. Construct Binary Tree from Inorder and Postorder Traversal 解题报告

题目：

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

Note:

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

这道题目和Leetcode 105十分类似，唯一的区别就在于后续遍历中，根节点处于最后一个位置。思路还是一样：首先记录下来每个value在中序遍历中的位置，然后递归地通过后序遍历序列来重构树。在重构的过程中，通过中序遍历的位置来确定左子树和右子树的大小，从而为下一步的递归计算出合适的参数。

代码：

/**
* 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) {
int len = inorder.size();
for (int i = 0; i < len; ++i) {
hash[inorder[i]] = i;
}
return rebuild(postorder, len - 1, 0, len);
}
private:
TreeNode* rebuild(vector<int>& postorder, int cur_pos, int start, int len) {
if (len <= 0) {
return NULL;
}
TreeNode* root = new TreeNode(postorder[cur_pos]);
int pos = hash[postorder[cur_pos]];
int len1 = pos - start;
int len2 = len - len1 - 1;
root->right = rebuild(postorder, cur_pos - 1, pos + 1, len2);
root->left = rebuild(postorder, cur_pos - len2 - 1, start, len1);
return root;
}
unordered_map<int, int> hash;
};