前序遍历和中序遍历树构造二叉树

        该问题用递归的思路很好解决，每一次取前序序列的首元素作为当前子树的根节点，然后在中序序列中找到对应的节点，以此可以确定根节点对应的左子树和右子树的序列长度，递归构造根节点的左子树和右子树即可。

TreeNode *execBuild(vector<int> &preorder, int prestart, int preend, vector<int> &inorder, int instart, int inend){
        
        TreeNode *result = new TreeNode(preorder[prestart]);
        
        if(prestart == preend){
            return result;
        }
        
        int i;
        for( i = instart; i <= inend; i++){
            if(inorder[i] == preorder[prestart])
                break;
        }
        
        if(i == instart){
            result -> left = nullptr;
        }else{
            result -> left = execBuild(preorder, prestart + 1, prestart + i - instart, inorder, instart, i - 1);
        }
        
        if(i == inend){
            result -> right = nullptr;
        }else{
            result -> right = execBuild(preorder, preend - (inend - i) + 1, preend, inorder, i + 1, inend);
        }
        return result;
    }

    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        // write your code here
        int size = preorder.size();
        if(size < 1)
            return nullptr;
        
        return execBuild(preorder, 0, size - 1, inorder, 0, size - 1);
    }

前序遍历和中序遍历树构造二叉树

原文：http://blog.csdn.net/ny_mg/article/details/48136549