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

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

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,0,inorder.size()-1,postorder,0,postorder.size()-1);
    }
    
    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=0;
        for(i=ileft;i<=iright;i++){
            if(inorder[i]==postorder[pright])
                break;
        }
        root->left=buildTree(inorder,ileft,i-1,postorder,pleft,pleft+i-ileft-1);
        root->right=buildTree(inorder,i+1,iright,postorder,pleft+i-ileft,pright-1);
        return root;
    }