LeetCode 145. 二叉树的后序遍历

给定一个二叉树，返回它的 后序 遍历。

示例:

输入: [1,null,2,3]  
   1
         2
    /
   3 

输出: [3,2,1]
进阶: 递归算法很简单，你可以通过迭代算法完成吗？

/**
 * 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> v;
 void dfs(TreeNode* root)
 {
     if (root == NULL) return;

     dfs(root->left);
     dfs(root->right);
     v.push_back(root->val);
 }

vector<int> postorderTraversal(TreeNode* root) {
    dfs(root);

    return v;
}
};

class Solution {
  public:
      map<TreeNode*, int> flag;
      stack< TreeNode*> st;
      vector<int> v;
      vector<int> postorderTraversal(TreeNode* root) {
         // memset(flag, 0, sizeof flag);
          if (root == NULL)return v;
          st.push(root);
          flag[root] = 1;
          while (st.size()) {
              TreeNode* p = st.top();
              int noNext = 1;

              if (p->right != NULL && flag[p->right ] == 0) {
                  st.push(p->right);
                  flag[p->right ] = 1;
                  noNext = 0;
              }
              if (p->left != NULL && flag[p->left ] == 0) {
                  st.push(p->left);
                  flag[p->left ] = 1;
                  noNext = 0;
              }

              if (noNext == 1) {
                  v.push_back(p->val);
                  st.pop();
              }
          }

          return v;
      }
  };