145. Binary Tree Postorder Traversal

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

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [3,2,1]

方法0: recursion

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> result;
        root = postorderHelper(root, result);
        return result;
    }
    
    TreeNode* postorderHelper(TreeNode* root, vector<int>& result){
        if (root == nullptr) return nullptr;
        postorderHelper(root->left, result);
        postorderHelper(root->right, result);
        result.push_back(root->val);
        return root;
    }
};

方法1: iterative

思路：

注意上图，postorder！=reverse（preorder）：左右顺序有变化。因此在preorder的基础上修改，将preorder入栈的左右序交换，最后用另一个栈反转一次。

易错点：

1. 左右顺序

Complexity

Time complexity : we visit each node exactly once, thus the time complexity is O(N), where NN is the number of nodes, i.e. the size of tree.

Space complexity : depending on the tree structure, we could keep up to the entire tree, therefore, the space complexity is O(N).

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> result;
        if (!root) return result;
        stack<TreeNode*> st1;
        stack<TreeNode*> st2;
        st1.push(root);
        while (!st1.empty()){
            auto top = st1.top();
            st1.pop();
            
            st2.push(top);
            if (top -> left) {
                st1.push(top -> left);
            }
            if (top -> right) {
                st1.push(top -> right);
            }
        }
        
        while (!st2.empty()){
            auto top = st2.top();
            st2.pop();
            
            result.push_back(top -> val);
        }
        return result;
    }
};