105. Construct Binary Tree from Preorder and Inorder Traversal

Posted 2020-10-10

Given preorder and inorder traversal of a tree, construct the binary tree.

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

方法一：迭代

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
13         if(preorder.size()==0)
14             return NULL;
15         stack<TreeNode*>s;
16         TreeNode*root,*t;
17         int flag=0,i=0,j=0;
18         root=new TreeNode(preorder[0]);
19         t=root;
20         s.push(root);
21         i++;
22         while(i<preorder.size())
23         {
24             if(!s.empty()&&s.top()->val==inorder[j])
25             {
26                 t=s.top();
27                 s.pop();
28                 flag=1;
29                 j++;
30             }
31             else
32             {
33                 if(flag==0)
34                 {
35                     t->left=new TreeNode(preorder[i]);
36                     t=t->left;
37                     s.push(t);
38                     i++;
39                 }
40                 else
41                 {
42                     flag=0;
43                     t->right=new TreeNode(preorder[i]);
44                     t=t->right;
45                     s.push(t);
46                     i++;
47                 }
48             }
49         }
50            return root;
51         }
52 };

方法二：递归