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.
recursive answer:
递归方法：在preorder找到root节点，接着在inorder里找到root节点，则inorder被分为两部分，一部分是left，一部分是right。在确定right支的时候要注意，在preorder里面确定right支的root节点，位置是left支节点个数和加上root节点。

class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
return myBT(0,preorder,0,inorder.size() - 1,inorder);
}

TreeNode * myBT(int preID, vector<int>& preorder, int inID, int inEnd, vector<int>& inorder){
if(inID < 0 || preID >= preorder.size() || inID > inEnd || inEnd >= inorder.size())
return NULL;
TreeNode * root = new TreeNode(preorder[preID]);
int i;
for(i = inID; i <= inEnd; i ++){
if(inorder[i] == preorder[preID])
break;
}
root->left = (myBT(preID + 1,preorder,inID,i - 1,inorder));
root->right = (myBT(preID + i - inID + 1,preorder,i + 1,inEnd,inorder));
return root;
}
};

iterative answer:
好像有点复杂。。。没空练习。。。