题目描写叙述：

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].

思路：以中右左的顺序完毕对树的遍历。然后再将结果反转。

代码：

vector
     
       Solution::postorderTraversal(TreeNode *root){    vector
      
        postorderSequence;    stack
       
         treeNodeStack;    TreeNode * node = root;    if(node == NULL)        return postorderSequence;    postorderSequence.push_back(node->val);    if(node->left)        treeNodeStack.push(node->left);    if(node->right)        treeNodeStack.push(node->right);    while(!treeNodeStack.empty())    {        node = treeNodeStack.top();        treeNodeStack.pop();        postorderSequence.push_back(node->val);        if(node->left)            treeNodeStack.push(node->left);        if(node->right)            treeNodeStack.push(node->right);    }    reverse(postorderSequence.begin(),postorderSequence.end());    return postorderSequence;}