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?
方法一:递归遍历
/**
 * 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> nums;
    vector<int> postorderTraversal(TreeNode* root) {
        postorder(root);
        return nums;
    }
    void postorder(TreeNode * root){
        if(root==NULL) return;
        if(root->left) postorder(root->left);
        if(root->right) postorder(root->right);
        nums.push_back(root->val);
    }
};Leetcode[145]-Binary Tree Postorder Traversal
原文:http://blog.csdn.net/dream_angel_z/article/details/46480869