Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},
1 \ 2 / 3
return [3,2,1].
/**
* 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:
vector<int> postorderTraversal(TreeNode* root) {
vector<int> path;
postOrder(root, path);
return path;
}
/*
void postOrder(TreeNode* root, vector<int> &path)
{
//遞歸寫法
if (root)
{
postOrder(root->left, path);
postOrder(root->right, path);
path.push_back(root->val);
}
}
*/
void postOrder(TreeNode* root, vector<int> &path)
{
//非遞歸寫法
stack<TreeNode *> TreeNodeStack;
TreeNode *plastvisit = NULL; //記錄結點是否訪問過
while (root != NULL || !TreeNodeStack.empty())
{
while (root != NULL)
{
TreeNodeStack.push(root);
root = root->left;
}
root = TreeNodeStack.top();
if (root->right == NULL || plastvisit == root->right)
{
path.push_back(root->val);
plastvisit = root;
TreeNodeStack.pop();
root = NULL;
}
else
root = root->right;
}
}
};