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# 106. Construct Binary Tree from Inorder and Postorder Traversal
## 1.問題
## 2.想法
* 提問
* 確認題意: root是NULL是true或false?
* function header, parameter
* r input
* 說明想法
* 想法與105題相同, 由於postorder序列建構tree, 必須由序列的最後往前
* 測試計算複雜度
## **3.程式碼**
```
/**
* 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& inorder, vector& postorder) {
int pos = postorder.size() - 1;
return buildBinary(inorder, postorder, 0, postorder.size() - 1, pos);
}
private:
TreeNode* buildBinary(vector& inorder, vector& postorder, int left, int right, int& pos) {
if (pos < 0 || left > right) {
return NULL;
}
int i = 0;
for (i = left; i <= right; i++) {
if (inorder[i] == postorder[pos]){
break;
}
}
TreeNode* node = new TreeNode(postorder[pos]);
pos--;
node->right = buildBinary(inorder, postorder, i + 1, right, pos);
node->left = buildBinary(inorder, postorder, left, i - 1, pos);
return node;
}
};
```