[leetcode] 145. Binary Tree Postorder Traversal 解题报告

题目链接：https://leetcode.com/problems/binary-tree-postorder-traversal/

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?

思路：递归的比较简单。迭代的想了好久没想到怎么做，后来看到居然是按照先序的思路，左右子树入栈顺序相反，最后将得到后序相反的序列，然后将这个序列翻转一下就可以了，感觉这种做法也挺意外的。180题了，很快就到２００题的关卡了。

两种代码如下：

递归：

/** * 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) {        if(!root) return result;        postorderTraversal(root->left);        postorderTraversal(root->right);        result.push_back(root->val);        return result;    }private:    vector<int> result;};

迭代：

/** * 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) {        if(!root) return result;        st.push(root);        while(!st.empty())        {            TreeNode* tem = st.top();            st.pop();            result.push_back(tem->val);            if(tem->left) st.push(tem->left);            if(tem->right) st.push(tem->right);        }        reverse(result.begin(), result.end());        return result;    }private:    stack<TreeNode*> st;    vector<int> result;};

第二种方法参考：https://leetcode.com/discuss/66062/c-iterative-solution-0ms-with-a-stack