数据结构 – 从有序和级别顺序遍历构造二叉树

2019年8月4日 224次阅读

首先,我想说明这不是一个功课.我正在准备面试并遇到这个问题.我想我们可以通过
in-order和
level-order遍历的定义. :-).

例如：

      50
   /      \
 10        60
/  \       /  \
5   20    55    70
        /     /  \
      51     65    80

The in-order and level-order traversal of the above tree are:

5, 10, 20, 50, 51, 55, 60, 65, 70, 80

50, 10, 60, 5, 20, 55, 70, 51, 65, 80

我的想法：

(1) traversal the level-order array to find out the first element
which appears in the in-order array. We call this element as current
root.

(2) find the index of current root in the in-order array. The in-order
array is separated by the index. The left side of the in-order array
is the left sub-tree of the current root and the right side of the
in-order array is the right sub-tree of the current root.

(3) update the in-order array as its left side and then go to step 1.

(4) update the in-order array as its right side and then go to step 2.

以上面的树为例.

06001

谁能帮我验证我的解决方案？
如果给另一个人真的很感激.

最佳答案 我认为你走在正确的轨道上.下面是我使用您的数据制定的工作代码.

/*
//construct a bst using inorder & levelorder traversals.
//inorder    - 5, 10, 20, 50, 51, 55, 60, 65, 70, 80
//levelorder - 50, 10, 60, 5, 20, 55, 70, 51, 65, 80
         50
      /      \
    10        60
   /  \       /  \
  5   20    55    70
            /     /  \
          51     65    80
 */
struct node *construct_bst3(int inorder[], int levelorder[], int in_start, int in_end)
{
    static int levelindex = 0;
    struct node *nnode = create_node(levelorder[levelindex++]);

    if (in_start == in_end)
        return nnode;

    //else find the index of this node in inorder array. left of it is left subtree, right of this index is right.
    int in_index = search_arr(inorder, in_start, in_end, nnode->data);

    //using in_index from inorder array, constructing left & right subtrees.
    nnode->left  = construct_bst3(inorder, levelorder, in_start, in_index-1);
    nnode->right = construct_bst3(inorder, levelorder, in_index+1, in_end);

    return nnode;
}
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