Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
题意:根据中序遍历和后序遍历,构建二叉树
思路很清晰,做法很简单,就不讲了。
一开始我写了一个递归的解法,本地测试数据都OK,无奈提交的时候内存超出限制,下面先给出超出内存的代码:
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 public:12 TreeNode* buildTree(vector & inorder, vector & postorder) {13 if(inorder.size()==0)14 return nullptr;15 if(inorder.size()==1)16 return new TreeNode(inorder[0]);17 int fath=postorder[postorder.size()-1];18 TreeNode* root=new TreeNode(fath);19 int flag=-1;20 for(int i=0;i