LeetCode 144 — Binary Tree Preorder Traversal（C++ Java Python）

题目：http://oj.leetcode.com/problems/binary-tree-preorder-traversal/

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?

题目翻译：

给定一个二叉树，返回其节点值的前序遍历。
例如：
给定二叉树{1,#,2,3}，
返回[1,2,3]。
注：递归的解法是简单的，你可以用迭代实现呢？
分析：

        使用stack来实现迭代。

C++实现1（递归版）：

/** * 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> result;    if(root != NULL)    {    result.push_back(root->val);    vector<int> left = preorderTraversal(root->left);    result.insert(result.end(), left.begin(), left.end());    vector<int> right = preorderTraversal(root->right);    result.insert(result.end(), right.begin(), right.end());    }    return result;    }};

C++实现2（迭代版）：

class Solution {public:    vector<int> preorderTraversal(TreeNode *root) {    vector<int> result;    if(root == NULL)    {    return result;    }    std::stack<TreeNode*> nodeStack;    nodeStack.push(root);    while(!nodeStack.empty())    {    TreeNode *node = nodeStack.top();    result.push_back(node->val);    nodeStack.pop();    if(node->right)    {    nodeStack.push(node->right);    }    if(node->left)    {    nodeStack.push(node->left);    }    }    return result;    }};

Java实现1（递归版）：

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public ArrayList<Integer> preorderTraversal(TreeNode root) {ArrayList<Integer> result = new ArrayList<Integer>();if (root != null) {result.add(root.val);result.addAll(preorderTraversal(root.left));result.addAll(preorderTraversal(root.right));}return result;    }}

Java实现2（迭代版）：

public class Solution {    public ArrayList<Integer> preorderTraversal(TreeNode root) {ArrayList<Integer> result = new ArrayList<Integer>();if (root == null) {return result;}Stack<TreeNode> nodeStack = new Stack<TreeNode>();nodeStack.push(root);while (!nodeStack.empty()) { TreeNode node = nodeStack.pop();result.add(node.val);if (node.right != null) {nodeStack.push(node.right);}if (node.left != null) {nodeStack.push(node.left);}}return result;    }}

Python实现1（递归版）：

# 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):        result = []                if root != None:            result.append(root.val)                        left = self.preorderTraversal(root.left)            result.extend(left)                        right = self.preorderTraversal(root.right)            result.extend(right)                return result

Python实现2（迭代版）：

class Solution:    # @param root, a tree node    # @return a list of integers    def preorderTraversal(self, root):        result = []                if root == None:            return result                nodeStack = []        nodeStack.append(root)                while len(nodeStack) != 0:            node = nodeStack.pop()            result.append(node.val)                        if node.right != None:                nodeStack.append(node.right)                        if node.left != None:                nodeStack.append(node.left)                    return result

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参考资料：http://www.geeksforgeeks.org/iterative-preorder-traversal/