【C++】【LeetCode】105. Construct Binary Tree from Preorder and Inorder Traversal
程序员文章站
2022-03-03 10:13:05
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题目
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
思路
前序排列,根结点在最前面;中序排列,根结点前面是左子树的结点,后面是右子数的结点。以此将左右子树分开,又可以按照前序排列继续分割。
代码
/**
* 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) {
TreeNode* root;
if (preorder.size() == 0) {
return NULL;
}
root = helper(root, preorder, 0, preorder.size() - 1, inorder, 0, preorder.size() - 1);
return root;
}
TreeNode* helper(TreeNode* root, vector<int>& preorder, int start1, int end1, vector<int>& inorder, int start2, int end2) {
if (end1 < start1) {
return root;
} else {
root = new TreeNode(preorder[start1]);
int i = 0;
for (; i < end2-start2; i++) {
if (inorder[i+start2] == preorder[start1]) {
break;
}
}
root->left = (i < 1) ? NULL : helper(root->left, preorder, start1 + 1, i+start1, inorder, start2, start2 + i - 1);
root->right = (i + start1 + 1 > end1) ? NULL : helper(root->right, preorder, i+start1 + 1, end1, inorder, i+start2 + 1, end2);
return root;
}
}
};
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