从前序中序遍历或中序后序遍历恢复二叉树

 TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        //剑指Offer 第2版 64页
        if(preorder.size() == 0 || inorder.size() == 0) return nullptr;
        return build(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1); 
    }
    
    TreeNode* build(vector<int>& preorder, int preorder_begin, int preorder_end, vector<int>& inorder, int inorder_begin, int inorder_end){
        //前序遍历第一个元素是根节点
        TreeNode* root = new TreeNode(preorder[preorder_begin]);
        
        //截止条件
        if(preorder_begin == preorder_end) return root;
        
        //根据根节点的值在中序遍历中划分出左子树和右子树
        int root_inorder = inorder_begin;
        for(; root_inorder <= inorder_end; root_inorder++){
            if(preorder[preorder_begin] == inorder[root_inorder]){
                break;
            }
        }
        //if(root_inorder == inorder.size()) //没有找到根节点
        int leftlength = root_inorder - inorder_begin;
        int leftpreorderend = preorder_begin + leftlength;
        //根据中序遍历的划分出的左右子树，更新preorder_begin,preorder_end,inorder_begin,inorder_end
        if(leftlength > 0){
            root->left = build(preorder,preorder_begin + 1,leftpreorderend, inorder, inorder_begin, root_inorder-1);
        }
        if(inorder_end - root_inorder > 0){
            root->right = build(preorder,leftpreorderend+1,preorder_end, inorder, root_inorder+1, inorder_end);
        }
        return root;
    }

TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        if(inorder.size() == 0 || postorder.size() == 0) return nullptr;
        return build(inorder,0,inorder.size()-1,postorder,0,postorder.size()-1);
    }
    TreeNode* build(vector<int>& inorder, int inorder_begin, int inorder_end, vector<int>& postorder, int postorder_begin, int postorder_end){
        if(inorder_begin > inorder_end || postorder_begin > postorder_end) return nullptr;
        
        //找到根节点
         TreeNode* root = new TreeNode(postorder[postorder_end]);
        
        //if(inorder_begin == inorder_end) return root;
        
        
       
        
        int root_pos = inorder_begin;
        for(;root_pos<=inorder_end;root_pos++){
            if(inorder[root_pos] == postorder[postorder_end])
            {
                break;
            }
        }
        int left_len = root_pos - inorder_begin;
        
        root->left = build(inorder, inorder_begin, inorder_begin + left_len - 1,postorder, postorder_begin, postorder_begin + left_len - 1);
        root->right = build(inorder, root_pos + 1, inorder_end, postorder, postorder_begin+left_len,postorder_end-1);
        return root;
    }