Preface
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As the title says
Given the root node root of a binary tree, return its preorder, inorder, and postorder traversals.
Example 1
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输入：root = [1,null,2,3]
输出：[1,2,3]
Example 2
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输入：root = []
输出：[]
Example 3
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输入：root = [1]
输出：[1]
Implementation Code
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struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
class Solution {
public:
void inorder(TreeNode *root, vector<int> &res) {
if (!root) {return;}
//Preorder
res.push_back(root->val);
preorder(root->left,res);
preorder(root->right,res);
//Inorder
inorder(root->left, res);
res.push_back(root->val);
inorder(root->right,res);
//Postorder
postorder(root->left, res);
postorder(root->right, res);
res.push_back(root->val);
}
vector<int> inorderTraversal(TreeNode* root) {
vector <int> ans;
inorder(root, ans);
return ans;
}
};
144. Binary Tree Preorder Traversal
94. Binary Tree Inorder Traversal
145. Binary Tree Postorder Traversal
Quoted from codetop