105. Construct Binary Tree from Preorder and Inorder Traversal

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

Note:

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

/**
* 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 Create(preorder, 0, preorder.size(),inorder,0, inorder.size());
}
TreeNode* Create(vector<int>& preorder, int pre_i,int pre_j,vector<int>& inorder,int in_i,int in_j)
{
if(pre_i>=pre_j||in_i>=in_j) return NULL;

int mid=preorder[pre_i];
auto mid_i=find(inorder.begin()+in_i,inorder.begin()+in_j,mid);
int dis=mid_i-inorder.begin()-in_i;

TreeNode* root=new TreeNode(mid);
root->left=Create(preorder, pre_i+1,pre_i+1+dis,inorder,in_i,in_i+1+dis);
root->right=Create(preorder, pre_i+1+dis,pre_j,inorder,in_i+1+dis,in_j);
return root;
}
};

O(n)(19ms) solution without recusion.

/**
* 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) {
TreeNode *root=NULL; stack<TreeNode *> MyData;
if(preorder.empty()) return root;
root = new TreeNode(preorder[0]);
MyData.push(root); int index = 0;
for(int i=1; i<=preorder.size(); i++) {
TreeNode *cur = MyData.top();
if((MyData.top()->val)!=inorder[index]) {
cur->left = new TreeNode(preorder[i]);
MyData.push(cur->left);
} else {
while(!MyData.empty() && ((MyData.top()->val)==inorder[index])) {
cur=MyData.top(); MyData.pop(); index++;
}
if(index<inorder.size()) {
cur->right = new TreeNode(preorder[i]);
MyData.push(cur->right);
}
}
}
return root;
}
};