LeetCode OJ：Binary Tree Postorder Traversal（后序遍历二叉树）

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?

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递归当然很容易实现拉，但是要求是不使用递归来实现，先贴一个递归的代码：

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        if(!root) return ret;
        tranverse(root);
        return ret;
    }

    void tranverse(TreeNode * root)
    {
        if(!root) return;
        tranverse(root->left);
        tranverse(root->right);
        ret.push_back(root->val);
    }
private:
    vector<int> ret;
};

下面是非递归实现的代码，用一个栈来保存数据：
注意左右子树的压栈顺序

/**
 * 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> ret;
        if(!root) return ret;
        stack<TreeNode *> s;
        s.push(root);
        while(!s.empty()){
            TreeNode * t = s.top();
            if(!t->left && !t->right){
                ret.push_back(t->val);
                s.pop();
                continue;
            }
            if(t->right){
                s.push(t->right);
                t->right = NULL;
            }
            if(t->left){
                s.push(t->left);
                t->left = NULL;
            }
        }
        return ret;
    }
};