根据一棵树的前序遍历与中序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。
例如,给出
前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树:
3
/ \
9 20
/ \
15 7
剑指offer原题,值得细细品味,特别经典的利用递归构造二叉树题目
/**
* 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) {
if(preorder.size()==0 || preorder.empty() || inorder.empty())
return NULL;
return SetUpTree(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
}
TreeNode* SetUpTree(vector<int>& preorder, int preStart, int preEnd,
vector<int>& inorder, int inStart, int inEnd){
TreeNode *root = new TreeNode(preorder[preStart]);
int pos = findInRoot(root->val, inorder, inStart, inEnd);
int leftSize = pos - inStart;
int rightSize = inEnd - pos;
if(leftSize > 0)
root->left = SetUpTree(preorder, preStart+1, preStart+leftSize, inorder, inStart, inStart+leftSize-1);
if(rightSize > 0)
root->right = SetUpTree(preorder,preEnd-rightSize+1, preEnd, inorder, inEnd-rightSize+1,inEnd);
return root;
}
private:
int findInRoot(int rootval, vector<int> &arr, int inStart, int inEnd){
for(int i=inStart;i <= inEnd;i++){
if(arr[i]==rootval)
return i;
}
return -1;
}
};