Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
You may assume that duplicates do not exist in the tree.
Solution:
Basic binary tree question, but need to take care of boundary of array index.
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/** | |
* 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* helper(vector<int>& preorder, long pre_start, long pre_end, | |
vector<int>& inorder, long in_start, long in_end) | |
{ | |
if(pre_start>pre_end || pre_start>=preorder.size() || pre_end<0){ | |
return NULL; | |
} | |
TreeNode* newNode=new TreeNode(preorder[pre_start]); | |
//find | |
long pos=in_start; | |
while(inorder[pos]!=preorder[pre_start]){pos++;} | |
long l_in_start=in_start; | |
long l_in_end=pos-1; | |
long r_in_start=pos+1; | |
long r_in_end=in_end; | |
long l_num=max(long(0), l_in_end-l_in_start+1); | |
long r_num=max(long(0), r_in_end-r_in_start+1); | |
long l_pre_start=l_num==0? INT_MAX:pre_start+1; | |
long l_pre_end=l_pre_start+l_num-1; | |
long r_pre_start=l_num==0? pre_start+1:l_pre_end+1; | |
r_pre_start=r_pre_start>pre_end? INT_MAX:r_pre_start; | |
long r_pre_end=r_pre_start+r_num-1; | |
TreeNode* lnode=helper(preorder, l_pre_start, l_pre_end, | |
inorder, l_in_start, l_in_end); | |
TreeNode* rnode=helper(preorder, r_pre_start, r_pre_end, | |
inorder, r_in_start, r_in_end); | |
newNode->left=lnode; | |
newNode->right=rnode; | |
return newNode; | |
} | |
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) { | |
TreeNode* root=helper(preorder, 0, preorder.size()-1, | |
inorder, 0, inorder.size()-1); | |
return root; | |
} | |
}; |
Round2 solution:
Following solution is shorter, though result in a Memory Limit Exceeded.
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class Solution { | |
public: | |
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) { | |
if(preorder.empty()){ | |
return NULL; | |
} | |
vector<int> pre_left; | |
vector<int> pre_right; | |
vector<int> in_left; | |
vector<int> in_right; | |
int j=-1; | |
for(int i=0; i<inorder.size(); i++){ | |
if(inorder[i]==preorder[0]) {j=i; continue;} | |
if(j==-1) | |
in_left.push_back(inorder[i]); | |
else | |
in_right.push_back(inorder[i]); | |
} | |
for(int i=1; i<preorder.size(); i++){ | |
if(i<=j) | |
pre_left.push_back(preorder[i]); | |
else | |
pre_right.push_back(preorder[i]); | |
} | |
TreeNode* node = new TreeNode (preorder[0]); | |
TreeNode* node_left = buildTree(pre_left, in_left); | |
TreeNode* node_right = buildTree(pre_right, in_right); | |
node->left = node_left; | |
node->right = node_right; | |
return node; | |
} | |
}; |
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