题目
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
思路：递归
同【剑指Offer】面试题07. 重建二叉树
关键在与正确定位左右子树范围。
代码
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
if (preorder.empty()) return nullptr;
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 (pend < pstart) return nullptr;
int val = preorder[pstart];
TreeNode* root = new TreeNode(val);
auto it = find(inorder.begin() + istart, inorder.begin() + (iend + 1), val);//注意在[istart, iend]范围内搜索
int lenLeft = it - find(inorder.begin() + istart, inorder.begin() + (iend + 1), inorder[istart]);
root->left = helper(preorder, pstart + 1, pstart + lenLeft, inorder, istart, istart + lenLeft - 1);
root->right = helper(preorder, pstart + lenLeft + 1, pend, inorder, istart + lenLeft + 1, iend);
return root;
}
};
另一种写法
修改求根节点索引。
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
if (preorder.empty()) return nullptr;
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 (pend < pstart) return nullptr;
int val = preorder[pstart];
TreeNode* root = new TreeNode(val);
auto it = find(inorder.begin() + istart, inorder.begin() + (iend + 1), val);
int index = it - inorder.begin();
int lenLeft = index - istart;
root->left = helper(preorder, pstart + 1, pstart + lenLeft, inorder, istart, index - 1);
root->right = helper(preorder, pstart + lenLeft + 1, pend, inorder, index + 1, iend);
return root;
}
};