【LeetCode】Binary Tree Postorder Traversal
使用迭代的版本：
#include <vector>
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
std::vector<int> postorderTraversal(TreeNode *root) {
vector<int> result;
traversal(result, root);
return result;
}
void traversal(std::vector<int>& result, TreeNode *node)
{
if (node == NULL) return;
traversal(result, node->left);
traversal(result, node->right);
result.push_back(node->val);
}
};
不使用迭代的版本（使用栈）：
#include <stack>
#include <vector>
#include <algorithm>
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
// 相当于倒着遍历一遍节点，postorder的输出顺序应该是 (left -> right -> node)
// 那我们现在就按照倒叙遍历这颗树并利用栈的性质进行输出 (node -> right -> left)
// 每次在遍历的时候都把栈顶元素出栈，相当于得到当前未遍历节点集合的最后一个节点
std::vector<int> postorderTraversal(TreeNode *root) {
std::vector<int> result;
std::stack<TreeNode *> _stack;
if (root == NULL) return result;
_stack.push(root);
while (!_stack.empty())
{
// 出栈当前节点值
TreeNode *cur = _stack.top(); _stack.pop();
result.push_back(cur->val);
// 入栈左子节点
TreeNode *left = cur->left;
if (left) _stack.push(left);
// 入栈右子节点
TreeNode *right = cur->right;
if (right) _stack.push(right);
}
std::reverse(result.begin(), result.end());
return result;
}
};
#endif // BINARY_TREE_POSTORDER_TRAVERSAL_H