Topic links: https://leetcode-cn.com/problems/binary-tree-postorder-traversal/
Title Description
Given a binary tree, it returns postorder traversal.
Example:
输入: [1,null,2,3]
1
\
2
/
3
输出: [3,2,1]
Advanced : recursive algorithm is very simple, you can do it through an iterative algorithm?
Thinking
Non-recursive
1 by reversing nonrecursive preorder
The simple non-recursive preorder traversal modify 根->右->左
the order of traversal, then the resulting inverted recursive sequence
obtained postorder traversal左->右->根
class Solution {
public:
vector<int> postorderTraversal(TreeNode* root) {
vector<int> ret;
if(!root) return ret;
stack<TreeNode *> s;
s.push(root);
while ( !s.empty()){
root = s.top();
s.pop();
ret.push_back(root->val);
if (root->left)
s.push(root->left);
if (root->right)
s.push(root->right);
}
reverse(ret.begin(),ret.end());
return ret;
}
};
2 stack storage unvisited right child node
With the preorder and inorder traversal, as with a Stack memory has not traverse the right subtree node.
Remember the last node traversed by a last, last access point if the child is the right, indicating that the right subtree has traversed before.
class Solution {
public:
vector<int> postorderTraversal(TreeNode* root) {
vector<int> ret;
if(!root) return ret;
stack<TreeNode *> s;
TreeNode *last = nullptr;
while (root || !s.empty()){
while (root){
s.push(root);
root = root->left;
}
root = s.top();
if(root->right == nullptr || root->right == last){
ret.push_back(root->val);
s.pop();
last = root;
root = nullptr;
}
else
root = root->right;
}
return ret;
}
};