LeetCode106 Construct Binary Tree from Inorder and Postorder Traversal

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

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

分析：

跟前一题方法一样，只需要分析出中序与后序恢复时，需从后序的最后一个元素读出来当做根节点，在中序中寻找其位置。

其余递归过程跟前序中序恢复一致。

代码：

1 /**
2  * Definition for a binary tree node.
3  * struct TreeNode {
4  *     int val;
5  *     TreeNode *left;
6  *     TreeNode *right;
7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8  * };
9  */
10 class Solution {
11 private:
12     TreeNode* helper(vector<int>& inorder, int inStart, int inEnd, vector<int>& postorder, int postStart, int postEnd) {
13         int sz = inEnd - inStart;
14         if (sz == 0) {
15             return nullptr;
16         }
17         int rootVal = postorder[postEnd - 1];
18         TreeNode* root = new TreeNode(rootVal);
19         int pos;
20         for (pos = 0; pos < sz; ++pos) {
21             if (inorder[inStart + pos] == rootVal) {
22                 break;
23             }
24         }
25         if (pos < 0) {
26             root -> left = nullptr;
27         }
28         else {
29             root -> left = helper(inorder, inStart , inStart + pos, postorder, postStart, postStart + pos);
30         }
31         if (pos + 1 >= sz) {
32             root -> right = nullptr;
33         }
34         else {
35             root -> right = helper(inorder, inStart + pos + 1, inEnd, postorder, postStart + pos, postEnd - 1);
36         }
37         return root;
38     }
39 public:
40     TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
41         return helper(inorder, 0, inorder.size(), postorder, 0, postorder.size());
42     }
43 };