Construct Binary Tree from Inorder and Postorder Traversal 中序和后序重建二叉树

Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.
Note:

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

/**
 * 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()==0||postorder.size()==0)
            return NULL;
        return solve(inorder,0,inorder.size()-1,postorder,0,postorder.size()-1);
    }
    
    TreeNode* solve(vector<int> &inorder ,int s1,int e1,vector<int> &postorder,int s2,int e2)
    {
        int rootVal=postorder[e2];
        int rootIndex=-1;
        for(int i=s1;i<=e1;i++)
            if(inorder[i]==rootVal)
            {
                rootIndex=i;
                break;
            }
        TreeNode *rootNode=new TreeNode(rootVal);
        int left=rootIndex-s1;
        int right=e1-rootIndex;
        if(left>0)
            rootNode->left=solve(inorder,s1,rootIndex-1,postorder,s2,s2+left-1);
        if(right>0)
            rootNode->right=solve(inorder,rootIndex+1,e1,postorder,s2+left,e2-1);
        return rootNode;
    }
};