Given a binary tree, return the postorder traversal of its nodes’ values.
For example:
Given binary tree {1,#,2,3},
return [3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
/*****************************************************************
* @Author : 楚兴
* @Date : 2015/2/24 01:27
* @Status : Accepted
* @Runtime : 4 ms
******************************************************************/
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL){}
};
class Solution {
public:
void helper(TreeNode *root, vector<int> &result) {
if (root)
{
helper(root->left, result);
helper(root->right, result);
result.push_back(root->val);
}
}
vector<int> postorderTraversal(TreeNode *root) {
vector<int> result;
helper(root, result);
return result;
}
};
class Solution {
public:
vector<int> postorderTraversal(TreeNode *root) {
vector<int> result;
if (root == NULL)
{
return result;
}
list<TreeNode*> nodes;
TreeNode *p = root;
TreeNode *cur;
nodes.push_back(p);
while(!nodes.empty())
{
cur = nodes.front();
//①当前结点不是叶子结点,但是孩子已经遍历过。
//②当前结点是叶子结点
if (cur->right == p || cur->left == p ||(cur->left == NULL && cur->right == NULL))
{
result.push_back(cur->val);
nodes.pop_front();
p = cur;
}
else
{
if (cur->right != NULL)
{
nodes.push_front(cur->right);
}
if (cur->left != NULL)
{
nodes.push_front(cur->left);
}
}
}
return result;
}
};
class Solution {
public:
vector<int> postorderTraversal(TreeNode *root) {
vector<int> result;
stack<TreeNode*> nodes;
TreeNode *p = NULL;
TreeNode *cur = root;
while(cur != NULL || !nodes.empty())
{
if (cur != NULL)
{
while (cur != NULL)
{
nodes.push(cur);
cur = cur->left;
}
}
else
{
cur = nodes.top();
if (cur->right == NULL || cur->right == p)
{
nodes.pop();
p = cur;
result.push_back(cur->val);
cur = NULL;
}
else
{
cur = cur->right;
}
}
}
return result;
}
};
先序遍历的实现见博文Binary Tree Preorder Traversal
[LeetCode] Binary Tree Postorder Traversal
原文:http://blog.csdn.net/foreverling/article/details/43922563