【二叉树经典问题】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.

做了第106题再来做这个就简单了，写递归的话思路一模一样：

/**
* 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 dfs(preorder,inorder,0,preorder.size()-1,0,inorder.size()-1);
}
TreeNode* dfs(vector<int>& preorder,vector<int>& inorder,int ps,int pe,int is,int ie){
if(ps>pe)
return nullptr;
TreeNode* root=new TreeNode(preorder[ps]);
int pos;
for(int i=0;i<=ie;i++){
if(inorder[i]==root->val){
pos=i;
break;
}
}
root->left=dfs(preorder,inorder,ps+1,pos-is+ps,is,pos-1);//pe-ps-1=pos-1-is => pe=pos-is+ps
root->right=dfs(preorder,inorder,pe-ie+pos+1,pe,pos+1,ie);//pe-ps=ie-pos-1 => ps=pe-ie+pos+1
return root;
}
};