LeetCode题解：Binary Tree Postorder Traversal-优快云博客

本文链接：https://blog.youkuaiyun.com/MagiSu/article/details/15024895

本文介绍了一种不使用递归实现二叉树后序遍历的方法，通过迭代方式利用双向链表实现节点的两次访问，最终达到与递归相同的效果。

Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

return [3,2,1].

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

思路：

无难度。

题解：

/**
 * Definition for binary tree
 * 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 == nullptr)
            return ret;
            
        list<pair<TreeNode*, bool>> traverse;
        traverse.push_back(make_pair(root, false));
        while(!traverse.empty())
        {
            auto& front = traverse.front();
            if (front.second)
            {
                // second time visit it
                ret.push_back(front.first->val);
                traverse.pop_front();
            }
            else
            {
                // first time visit it
                front.second = true;
                
                if (front.first->right)
                    traverse.push_front(make_pair(front.first->right, false));
                
                if (front.first->left)
                    traverse.push_front(make_pair(front.first->left, false));
            }
        }
        return ret;
    }
};