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.

For example, given

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2

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

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5

    3
   / \
  9  20
    /  \
   15   7

实现：

  • 递归方式，根据二叉树中序遍历和后续遍历，创建二叉树。中序遍历： 左-根-右， 后续遍历： 左-右-根， 所以可以通过后续遍历的最后一个节点找到根节点，并生成根节点， 因为二叉树没有重复节点， 就 可以在中序遍历数据中找到根节点，再分成左右两个子树，递归执行下去，遍历完所有节点。 通过postorder获取根节点， 通过inorder生成左右子树，查找节点在中序遍历数据组里位置，可以通过unordered_map实现。 
  • 不使用map情况下   时间复杂度 O(N^2)    , 使用map查找O(N)      
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        unordered_map <int, int> treemap;
        for (int i=0; i<inorder.size();i++) {
            treemap[inorder[i]] = i;
       }
        int index = inorder.size()-1;
        return  helper(inorder, postorder, 0, inorder.size()-1, &index, treemap);
    }
    TreeNode *helper(vector<int> & inorder, vector<int>& postorder, int left, int right, 
                     int *index , unordered_map<int, int>&treemap){
        //check
        if (left > right){
            return NULL;
        }
      
        // create root  node 
        /* destory postorder vector
        int val = postorder.back();
        TreeNode *node = new TreeNode(val);
        postorder.pop_back();
        */
        int val = postorder[*index];
        TreeNode *node = new TreeNode(val);
        (*index)--;
        
        
        // search inorder by array
        /* 
        int root = -1;
        for (int i=left ; i <=right; i ++){
            if (val == inorder[i]){
                root = i;
                break;
            }
        }
        if (root == -1){
            return node;
        }*/
        // search inorder by map
        int root = treemap[val];
    
        // create left & right node 
        node->right = helper(inorder, postorder,  root+1, right, index,  treemap);
        node->left  = helper(inorder, postorder, left, root-1, index, treemap);
       
     
        return node;
    }
};