LeetCode | Binary Tree Inorder Traversal
1337 ワード
タイトル:
Given a binary tree, return the inorder traversal of its nodes' values.
For example: Given binary tree {1,#,2,3}
, 1
\
2
/
3
return [1,3,2]
.
Note: Recursive solution is trivial, could you do it iteratively?
confused what "{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.
考え方:の古典的な中順序ソートアルゴリズムは、再帰的に実装されてもよいし、重み付きスタックで実装されてもよい.
コード: /**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode *root) {
if(root == NULL){
return vector<int>();
}
vector<int> left = inorderTraversal(root->left);
left.push_back(root->val);
vector<int> right = inorderTraversal(root->right);
for(int i = 0; i < right.size(); i++){
left.push_back(right[i]);
}
return left;
}
};
1
\
2
/
3
コード: /**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode *root) {
if(root == NULL){
return vector<int>();
}
vector<int> left = inorderTraversal(root->left);
left.push_back(root->val);
vector<int> right = inorderTraversal(root->right);
for(int i = 0; i < right.size(); i++){
left.push_back(right[i]);
}
return left;
}
};
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode *root) {
if(root == NULL){
return vector<int>();
}
vector<int> left = inorderTraversal(root->left);
left.push_back(root->val);
vector<int> right = inorderTraversal(root->right);
for(int i = 0; i < right.size(); i++){
left.push_back(right[i]);
}
return left;
}
};