144. Binary Tree Preorder Traversal (easy)

Given the root of a binary tree, return the preorder traversal of its nodes' values.

 

Example 1:

Input: root = [1,null,2,3]

Output: [1,2,3]

Explanation:

Example 2:

Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]

Output: [1,2,4,5,6,7,3,8,9]

Explanation:

Example 3:

Input: root = []

Output: []

Example 4:

Input: root = [1]

Output: [1]

 

Constraints:

• The number of nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

 

Follow up: Recursive solution is trivial, could you do it iteratively?

A:
--------------------   Recursive way. ------------

/**

* Definition for a binary tree node.

* 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:

vector<int> preorderTraversal(TreeNode* root) {

vector<int> res;

helper(root, res);

return res;

}

private:

void helper(TreeNode* root, vector<int>& res) {

if (!root)

return;

res.push_back(root->val);

helper(root->left, res);

helper(root->right, res);

}

};

-----------------Iterative way ---------------

simply use a stack to push right and left, and print the current node

/**
 * Definition for a binary tree node.
 * 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:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> S;
        if (root)
            S.push(root);
        while (!S.empty()) {
            auto tmp = S.top();
            S.pop();
            res.push_back(tmp->val);
            if (tmp->right) {
                S.push(tmp->right);
            }
            if (tmp->left) {
                S.push(tmp->left);
            }
        }
        return res;
    }
};

Mistakes:
1:  要注意， 先往stack里加右子树，再加左子树。
2:   这次忘了先check root == nullptr