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]

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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
    if(root == NULL) return res;

    TreeNode *p = root;
    stack<TreeNode *> sta;
    TreeNode *last = root;
    sta.push(p);
    while (!sta.empty())
    {
        p = sta.top();
        if( (p->left == NULL && p->right == NULL) || (p->right == NULL && last == p->left) || (last == p->right) )
        {
            res.push_back(p->val);
            last = p;
            sta.pop();
        }
        else 
        {
            if(p->right)
                sta.push(p->right);
            if(p->left)
                sta.push(p->left);
        }

    }
    return res;
  }
};

转载：https://www.cnblogs.com/rain-lei/p/3705680.html