【LeetCode]Binary Tree Preorder Traversal

標籤：leetcode   前序走訪   

    題意：

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?

    思路：

    寫個非遞迴的前序走訪，用 stack.

    代碼：    C++：

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

     Python：（python 寫棧也是可以的）

# Definition for a  binary tree node# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution:    # @param root, a tree node    # @return a list of integers    def preorderTraversal(self, root):        ans = []        s = []        while root != None or s:            while root != None:                ans.append(root.val)                s.append(root)                root = root.left            if s:                root = s[-1]                s.pop()                root = root.right        return ans

【LeetCode]Binary Tree Preorder Traversal