题意:根据二叉树的中序遍历和后序遍历恢复二叉树。

解题思路:看到树首先想到要用递归来解题。以这道题为例:如果一颗二叉树为{1,2,3,4,5,6,7},则中序遍历为{4,2,5,1,6,3,7},后序遍历为{4,5,2,6,7,3,1},我们可以反推回去。由于后序遍历的最后一个节点就是树的根。也就是root=1,然后我们在中序遍历中搜索1,可以看到中序遍历的第四个数是1,也就是root。根据中序遍历的定义,1左边的数{4,2,5}就是左子树的中序遍历,1右边的数{6,3,7}就是右子树的中序遍历。而对于后序遍历来讲,一定是先后序遍历完左子树,再后序遍历完右子树,最后遍历根。于是可以推出:{4,5,2}就是左子树的后序遍历,{6,3,7}就是右子树的后序遍历。而我们已经知道{4,2,5}就是左子树的中序遍历,{6,3,7}就是右子树的中序遍历。再进行递归就可以解决问题了。

 /**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
if(inorder.size()!=postorder.size()||inorder.size()<)
return NULL;
return build(inorder,postorder,,inorder.size()-,,postorder.size()-);
}
TreeNode *build(vector<int> &inorder, vector<int> &postorder, int startin,int endin,int startpost,int endpost){
if(startin>endin||startpost>endpost) return NULL;
if(startin==endin) return new TreeNode(inorder[startin]);
TreeNode *res = new TreeNode(postorder[endpost]);
int tmp=postorder[endpost];
int index=;
while((startin+index)<=endin){
if(inorder[startin+index]==tmp)
break;
index++;
}
res->left=build(inorder,postorder,startin,startin+index-,startpost,startpost+index-);
res->right=build(inorder,postorder,startin+index+,endin,startpost+index,endpost-);
return res;
}
};

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