LeetCode---(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.

第一个元素即a是根节点，从对应的中序中找到a。从而进一步知道其左边的b在左树中，其右边的c在右树中，这样结合前序递归可以还原出整个树。

/**
* 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* help(vector<int>& preorder,vector<int>& inorder,int fromp,int fromi,int length)
{
if(length==0)
return 0;
TreeNode* root=new TreeNode(preorder[fromp]);
int i=0;
for(i=fromi;inorder[i]!=preorder[fromp];)
i++;
root->left=help(preorder,inorder,fromp+1,fromi,i-fromi);
root->right=help(preorder,inorder,fromp+1+i-fromi,i+1,length-1-i+fromi);
return root;
}
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
return help(preorder,inorder,0,0,preorder.size());
}
};