题解 | #二叉树的中序遍历#
二叉树的中序遍历
https://www.nowcoder.com/practice/0bf071c135e64ee2a027783b80bf781d
递归
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
*
* @param root TreeNode类
* @return int整型vector
*/
vector<int> ans;
vector<int> inorderTraversal(TreeNode* root) {
// write code here
dfs(root);
return ans;
}
void dfs(TreeNode* root) {
if (!root) return;
dfs(root->left);
ans.push_back(root->val);
dfs(root->right);
}
};
迭代
/**
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* };
*/
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
*
* @param root TreeNode类
* @return int整型vector
*/
vector<int> inorderTraversal(TreeNode* root) {
// write code here
vector<int> res;
stack<TreeNode*> stk;
while(root || stk.size()) {
while(root) {
stk.push(root);
root = root->left;
}
root = stk.top();
stk.pop();
res.push_back(root->val);
root = root->right;
}
return res;
}
};
#二叉树的中序遍历#