[LeetCode] Binary Tree Postorder Traversal dfs,深度搜索

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?

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  一题后续遍历树的问题,很基础,统计哪里的4ms 怎么实现的。- -
 
#include <iostream>
#include <vector>
using namespace std;

/**
 * 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==NULL)  return ret;
        help_f(root,ret);
        return ret;
    }
    void help_f(TreeNode *node,vector<int> &ret)
    {
        if(node==NULL)  return;
        help_f(node->left,ret);
        help_f(node->right,ret);
        ret.push_back(node->val);
    }
};

int main()
{
    return 0;
}
原文地址:https://www.cnblogs.com/Azhu/p/4210779.html