【leetcode】144. Binary Tree Preorder Traversal

一、题目描述

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

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

   1    \     2    /   3

return [1,2,3].

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

题目解读：返回二叉树的前序遍历

方法一：递归

c++代码（0ms，23.98%）

/** * 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> result;    vector<int> preorderTraversal(TreeNode* root) {        if(root == NULL)    return result;        preorder(root);        return result;    }    void preorder(TreeNode* root){        if(root == NULL)    return;        result.push_back(root->val);        preorder(root->left);        preorder(root->right);    }//};

方法二：迭代

c++代码（0ms，23.98%）

/** * 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> preorderTraversal(TreeNode* root) {        stack<TreeNode*> q;        vector<int> result;        if(root == NULL)            return result;        TreeNode* p=root;        while(p!=NULL || !q.empty()){            while(p!=NULL){                result.push_back(p->val);                q.push(p);                p=p->left;            }//while            if(!q.empty()){                p=q.top()->right;                q.pop();            }        }        return result;    }};