105. Construct Binary Tree from Preorder and Inorder Traversal
Medium
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Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
preorder = [3,9,20,15,7] inorder = [9,3,15,20,7]
Return the following binary tree:
3 / \ 9 20 / \ 15 7
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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* buildTree(vector<int>& preorder, vector<int>& inorder) {
TreeNode* root=build(0,inorder.size()-1,0,preorder.size()-1,preorder,inorder);
return root;
}
TreeNode* build(int inl,int inr,int prel,int prer,vector<int>& preorder, vector<int>& inorder){
if(prel>prer) return NULL;
TreeNode* root=new TreeNode(preorder[prel]);
//root->val=preorder[prel];
//root->left=NULL;
//root->right=NULL;
if(inl==inr) return root;
int k;
for(k=inl;k<=inr;k++){//根据前序第一个为根找到中序中的根节点
if(preorder[prel]==inorder[k]){
break;
}
}
root->left=build(inl,k-1,prel+1,prel+(k-inl),preorder,inorder);//递归建立左子树
root->right=build(k+1,inr,prel+(k-inl)+1,prer,preorder,inorder);
return root;
}
};