題目描述

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

輸入

[2,1,3],[2,3,1]

輸出

{1,2,3}




//不需要輔助函式，簡單易懂
//後序遍歷容器的最後一個數是根節點，中序遍歷的根節點左邊是左子樹，右邊是右子樹，
 
//後序遍歷左子樹節點值相鄰，右子樹節點值也相鄰。由後序遍歷最後一個值將中序遍歷分成
//左右兩部分，再由這兩部分的size將後序遍歷分成左右兩部分，遞迴即可

class Solution {
public:
 TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
 if(inorder.empty())
 return NULL;
 int nodeval=postorder[postorder.size()-1];
 TreeNode* head=new TreeNode(nodeval);
 vector<int>::iterator iter=find(inorder.begin(),inorder.end(),nodeval);
 
 vector<int>leftin=vector<int>(inorder.begin(),iter);
 vector<int>rightin=vector<int>(iter+1,inorder.end());
 int left=leftin.size();
 int right=rightin.size();
 if(left>0)
 {
 vector<int>leftpo=vector<int>(postorder.begin(),postorder.begin()+left);
 head->left=buildTree(leftin,leftpo);
 
 }
 if(right>0)
 {
 vector<int>rightpo=vector<int>(postorder.begin()+left,postorder.begin()+left+right);
 head->right=buildTree(rightin,rightpo);
 }
 return head;
 }
};