binary-tree-postorder-traversal（后序遍历二叉树）

来源：互联网发布：做淘宝客服工资多少编辑：程序博客网时间：2024/06/06 21:38

题目描述

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].

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

非递归实现：

语言：C++ 运行时间： <1 ms 占用内存：8568K 状态：答案正确

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/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
classSolution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> v;
        stack<TreeNode*> s;
        map<TreeNode*,int>mp;//构造  节点指针，访问flag 的map
         
        while(root != NULL || !s.empty()){
            while(root != NULL){
                s.push(root);
                mp[root] = 1;       //怎么记录flag？？？？
                root = root->left;
            }
            while(!s.empty() && mp[s.top()]==2){
                v.push_back(s.top()->val);
                s.pop();
            }
            if(!s.empty()){
                mp[s.top()] = 2;
                root = s.top()->right;
            }
        }
        returnv;
    }
};

/**

 * Definition for binary tree

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

*/

//=====================递归方法istrivial。不过还是写出来吧=================================

classSolution {

public:

    voidpostOrder(TreeNode *root,vector<int>&vec){

        if(root != NULL){

            postOrder(root->left,vec);

            postOrder(root->right,vec);

            vec.push_back(root->val);

}

}

    vector<int> postorderTraversal(TreeNode *root) {

        vector<int>vec;

        postOrder(root,vec);

        returnvec;      

}

};

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