LeetCode 106. Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树 C++

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

inorder = [,,,,]
postorder = [,,,,]

Return the following binary tree:

   / \

    /  \
      

中序、后序遍历得到二叉树，可以知道每一次新数组的最后一个数为当时子树的根节点，每次根据中序遍历的根节点的左右两边确定左右子树，再对应后序的左右子树，不停递归得到根节点，可以建立二叉树。每次由循环得到根节点在中序数组中坐标i

由中序遍历知：每一次inorder的左子树范围[ileft，i-1],右子树范围[i+1,iright]

由后序遍历知：每一次postorder的左子树范围[pleft,pleft+i-ileft-1],右子树范围[pleft+i-ileft,pright-1]。C++

 TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
         return buildTree(inorder,,inorder.size()-,postorder,,postorder.size()-);
     }

     TreeNode* buildTree(vector<int>& inorder,int ileft,int iright,vector<int>& postorder,int pleft,int pright){
         if(ileft>iright||pleft>pright)
             return NULL;
         TreeNode* root=new TreeNode(postorder[pright]);
         int i=;
         for(i=ileft;i<=iright;i++){
             if(inorder[i]==postorder[pright])
                 break;
         }
         root->left=buildTree(inorder,ileft,i-,postorder,pleft,pleft+i-ileft-);
         root->right=buildTree(inorder,i+,iright,postorder,pleft+i-ileft,pright-);
         return root;
     }