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+# [106. Construct Binary Tree from Inorder and Postorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/)
+# 思路
+根据中序遍历和后序遍历构建二叉树。       
+其实和105题（根据中序遍历和前序遍历构建二叉树）是类似的，由于后序遍历的最后一个肯定是根，所以原二叉树的根节点可以知道，然后由于树中没有相同元素，
+所以我们可以在中序遍历中也定位出根节点的位置，并以根节点的位置将中序遍历拆分为左右两个部分，分别对其递归调用原函数。
+其中在中序遍历中定位根节点可以通过hash（即unordered_map）实现。
+
+# C++
+``` C++
+class Solution {
+private:
+    TreeNode* helper(int num, vector<int>& postorder, int post_last, 
+        vector<int>& inorder, int in_start, 
+        unordered_map<int, int>& mp){
+        if(num == 0) return NULL;
+        int val = postorder[post_last];
+        TreeNode *node = new TreeNode(val);
+        int i = mp[val] - in_start; // i代表左子树有多少个节点
+        node -> left = helper(i, postorder, post_last - num + i, inorder, in_start, mp);
+        node -> right = helper(num-i-1, postorder, post_last-1, inorder, 
+                               in_start+i+1, mp);
+        return node;
+    }
+public:
+    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
+        unordered_map<int, int>mp;
+        // 用mp记录某个节点在中序遍历数组中的索引
+        for(int i = 0; i < inorder.size(); i++) mp[inorder[i]] = i;
+        int num = postorder.size();
+        return helper(num, postorder, num - 1, inorder, 0, mp); 
+        
+    }
+};
+```
+