145. Binary Tree Postorder Traversal 二叉树后序遍历

题目链接
tag:

• Hard；

question：
  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?

思路：
  那么在94. Binary Tree Inorder Traversal 二叉树中序遍历中的解法三也可以改动一下变成后序遍历，代码如下：

/**
 * 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 postorderTraversal(TreeNode* root) {
        vector res;
        stack s;
        TreeNode *p = root;
        while (!s.empty() || p) {
            if (p) {
                s.push(p);
                res.insert(res.begin(), p->val);
                p = p->right;
            } else {
                TreeNode *t = s.top(); s.pop();
                p = t->left;
            }
        }
        return res;
    }
};