題目：
Given the root of a binary tree, return the postorder traversal of its nodes' values.
Example 1:
Input: root = [1,null,2,3] Output: [3,2,1]
Example 2:
Input: root = [] Output: []
Example 3:
Input: root = [1] Output: [1]
Example 4:
Input: root = [1,2] Output: [2,1]
Example 5:
Input: root = [1,null,2] Output: [2,1]
Constraints:
- The number of the 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?
代碼：
/**
* 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:
void backPostorder(TreeNode* root, vector<int>& res) {
if(root == NULL)
return;
backPostorder(root->left, res);
backPostorder(root->right, res);
res.push_back(root->val);
return;
}
vector<int> postorderTraversal(TreeNode* root) {
vector<int> res;
if(root == NULL)
return res;
backPostorder(root, res);
return res;
}
};
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