leetcode_construct_binary_tree_from_preorder_and_inorder_traversal

发表于

难度：Medium

解题思路：

      a
  b     c
d  e  f  g

前序遍历：a b d e c f g    
中序遍历：d b e a f c g

可知：a一定是root，在中序遍历中，a的左侧即左子树，a的右侧为右子树，并可以分别计算出各自的长度。计算出长度后，可以在前序遍历中，找出左子树的前序遍历，右子树的前序遍历，进行递归。

代码如下：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return buildSubTree(preorder,inorder,0,preorder.size()-1,0,inorder.size()-1);
    }
    TreeNode* buildSubTree(vector<int> &preorder, vector<int> &inorder, int pre_left, int pre_right, int in_left, int in_right)
    {
        if(pre_left > pre_right || in_left > in_right)
            return NULL;
        else if(pre_left == pre_right || in_left == in_right)
        {
            TreeNode *new_node = new TreeNode(preorder[pre_left]);
            return new_node;
        }
        else
        {
            TreeNode *new_node = new TreeNode(preorder[pre_left]);
            int i = in_left;
            for(; i <= in_right; i++)
            {
                if(inorder[i] == preorder[pre_left])
                    break;
            }
            int new_node_position_in_order = i;
            int left_length = new_node_position_in_order-in_left;
            new_node->left = buildSubTree(preorder, inorder, pre_left+1, pre_left+left_length, in_left, new_node_position_in_order-1);
            new_node->right = buildSubTree(preorder, inorder, pre_left+left_length+1, pre_right, new_node_position_in_order+1,in_right);
            return new_node;
        }
    }
};

运行结果：43ms，超过39.61%