Given preorder and inorder traversal of a tree, construct the binary tree.
根据树的前序遍历和中序遍历,构建二叉树
/**
* 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 help(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);
}
TreeNode* help(vector<int>& preorder,int begin1,int end1,vector<int>& inorder,int begin2,int end2)
{
if(begin1>end1)
return NULL;
else if(begin1==end1)
return new TreeNode(preorder[begin1]);
TreeNode* root=new TreeNode(preorder[begin1]);
int i;
for(i=begin2;i<end2;i++)
if(inorder[i]==preorder[begin1])
break;
int leftlen=i-begin2;
//左右子树怎么分的举个例子就明白了
root->left=help(preorder,begin1+1,begin1+leftlen,inorder,begin2,begin2+leftlen-1);
root->right=help(preorder,begin1+leftlen+1,end1,inorder,begin2+leftlen+1,end2);
return root;
}
};
leetcode No105. Construct Binary Tree from Preorder and Inorder Traversal
原文:http://blog.csdn.net/u011391629/article/details/52279387