145. Binary Tree Postorder Traversal -----------E

Given the root of a binary tree, return the postorder traversal of its nodes' values.

Follow up: Recursive solution is trivial, could you do it iteratively?

 

Example 1:

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]

Explanation:

Example 2:

Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]

Output: [4,6,7,5,2,9,8,3,1]

Explanation:

Example 3:

Input: root = []

Output: []

Example 4:

Input: root = [1]

Output: [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?
A:
-----------------递归解法-----------

/**

* Definition for a binary tree node.

* struct TreeNode {

* int val;

* TreeNode *left;

* TreeNode *right;

* TreeNode() : val(0), left(nullptr), right(nullptr) {}

* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}

* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}

* };

*/

class Solution {

public:

vector<int> postorderTraversal(TreeNode* root) {

vector<int> res;

helper(root, res);

return res;

}

private:

void helper(TreeNode* root, vector<int>& res) {

if (!root)

return;

helper(root->left, res);

helper(root->right, res);

res.push_back(root->val);

}

};

不论是用两个stack （还是每次加到前面，最后reverse回来）原理都是一样的。 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> S;
        if(root)
            S.push(root);
        while(!S.empty()){
            auto tmp = S.top();
            S.pop();
            res.push_back(tmp->val);
            if(tmp->left)
                S.push(tmp->left);
            if(tmp->right)
                S.push(tmp->right);
        }
        reverse(res.begin(), res.end());
        return res;
    }
};

Mistakes:
1： 忘了先check root为null 的情况
--------------------------第三遍------（用了一个stack,一个set（set来记录是否被visited过））--------------------
2：  把if写成了while   ，  哎，丢死人了
3：  写成了先  压栈右边，再压栈左边的了。     这个， 当时是考虑到正常顺序，是要先遍历左子树，再遍历右子树的。   但是， 但是，没有考虑到， 其实，我们是reversely 加入到al的最前面的。 因此，  要先遍历右子树的。

public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
         List<Integer> result = new LinkedList();
         if(root ==null)
            return result;
         Stack<TreeNode> stack = new Stack();
         Set<TreeNode> set = new HashSet();
         stack.push(root);
         set.add(root);
         while(!stack.isEmpty()){
             TreeNode node = stack.pop();
             if(set.contains(node)){// first time to find this node
                stack.push(node);
                if(node.right!=null){
                    stack.push(node.right);
                    set.add(node.right);
                }
                if(node.left!=null){
                    stack.push(node.left);
                    set.add(node.left);
                }
                 set.remove(node);
             }else{// node should be printed directly
                 result.add(node.val);
             }
         }// end of while
         return result;
    }
}

Learned: