- 题目:根据二叉树的前序和后序遍历结果,重新构造二叉树
- 难度:Medium
- 思路:遍历前序结果,找到根节点在中序遍历中的index,然后继续构造左右子树。【确定左右子树的元素所在区间,包括前序区间和中序区间】
- 代码:
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
map<int, int> m;
if(preorder.empty()){
return NULL;
}
int inlen = inorder.size();
int prelen = preorder.size();
if(inlen == 0 || prelen == 0 || inlen != prelen){
return NULL;
}
return helper(preorder, 0, inlen-1, inorder, 0, inlen-1);
}
TreeNode* helper(vector<int>& preorder, int ps, int pe, vector<int>& inorder, int is, int ie){
if(ps > pe || is > ie){
return NULL;
}
TreeNode* root = new TreeNode(preorder[ps]);
auto f = find(inorder.begin() + is, inorder.begin() + ie, preorder[ps]);
int ir = distance(inorder.begin(), f);
root->left = helper(preorder, ps+1, ps+ir-is, inorder, is, ir-1);
root->right = helper(preorder, ps+ir-is+1, pe, inorder, ir+1, ie);
return root;
}
};