LeetCode Binary Tree PostorderTranversal

Problem Description

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].

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

Problem Solution

1. 递归方案

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    vector<int> nodeVec;
public:
    void traverse(TreeNode *root){
        if(root==NULL)
            return;
        traverse(root->left);
        traverse(root->right);
        nodeVec.push_back(root->val);
    }
    vector<int> postorderTraversal(TreeNode *root) {
        traverse(root);
        return nodeVec;
    }
};

2. 非递归方案

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    vector<int> nodeVec;
public:void iterTraverse(TreeNode *root){
        if(root==NULL)
            return;
        stack<TreeNode*> st;
        
        TreeNode *pCur,*pPrev=NULL; //pCur: current tree node, pPrev: previous visited tree node
        st.push(root);
        while(!st.empty())
        {
            pCur=st.top();
            if((pCur->left == NULL && pCur->right == NULL) || (pPrev != NULL && (pCur->left==pPrev || pCur->right==pPrev)))
            {
                nodeVec.push_back(pCur->val);                
                pPrev=pCur;
                st.pop();
            }
            else
            {
                if(pCur->right != NULL)
                    st.push(pCur->right);
                if(pCur->left != NULL)
                    st.push(pCur->left);
            }
        }
    }
    vector<int> postorderTraversal(TreeNode *root) {
        iterTraverse(root);
        return nodeVec;
    }
};