Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
以前写traversal的时候其实是死记硬背。。。谁在前谁在后,遇到这个题才这要搞懂。youtube有个人一步步给画了。不贴了
假设树 pre order为 a b c d e f g, 那么 a为root. 如果他的inorder 为 b c d a e f g 那么 bcd为左,def为右,然后递归的对左右做同样的操作。
/**
* Definition for binary tree
* 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 helper(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
}
TreeNode* helper(vector<int>& preorder, int pStart, int pEnd, vector<int>& inorder, int iStart, int iEnd){
if (pStart>pEnd)
return NULL;
if (pStart==pEnd)
return new TreeNode(preorder[pStart]);
int rootInd;
for (int i=iStart; i<=iEnd; i++){
if (inorder[i]==preorder[pStart]){
rootInd=i;
break;
}
}
TreeNode* root=new TreeNode(preorder[pStart]);
int leftLen=rootInd-iStart;
root->left=helper(preorder, pStart+1, pStart+leftLen, inorder, iStart, rootInd-1);
root->right=helper(preorder, pStart+leftLen+1, pEnd, inorder, rootInd+1, iEnd);
return root;
}
};