LeetCode 102. Binary Tree Level Order Traversal--递归，迭代-Python，Java解法

题目地址：

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3/ \9  20/  \15   7

return its level order traversal as:

[[3],[9,20],[15,7]
]

这道题可以用迭代做，也可以用递归做。
Python递归解法（比较简单）：

class Solution:def levelOrder(self, root):""":type root: TreeNode:rtype: List[List[int]]"""levels = []if not root:return levelsdef helper(node, level):# start the current levelif len(levels) == level:levels.append([])# append the current node valuelevels[level].append(node.val)# process child nodes for the next levelif node.left:helper(node.left, level + 1)if node.right:helper(node.right, level + 1)helper(root, 0)return levels

Python迭代解法,使用队列

from collections import deque
class Solution:def levelOrder(self, root):""":type root: TreeNode:rtype: List[List[int]]"""levels = []if not root:return levelslevel = 0queue = deque([root,])while queue:# start the current levellevels.append([])# number of elements in the current level level_length = len(queue)for i in range(level_length):node = queue.popleft()# fulfill the current levellevels[level].append(node.val)# add child nodes of the current level# in the queue for the next levelif node.left:queue.append(node.left)if node.right:queue.append(node.right)# go to next levellevel += 1return levels

java递归解法：

class Solution {List<List<Integer>> levels = new ArrayList<List<Integer>>();public void helper(TreeNode node, int level) {// start the current levelif (levels.size() == level)levels.add(new ArrayList<Integer>());// fulfil the current levellevels.get(level).add(node.val);// process child nodes for the next levelif (node.left != null)helper(node.left, level + 1);if (node.right != null)helper(node.right, level + 1);}public List<List<Integer>> levelOrder(TreeNode root) {if (root == null) return levels;helper(root, 0);return levels;}
}

java迭代解法：

class Solution {public List<List<Integer>> levelOrder(TreeNode root) {List<List<Integer>> levels = new ArrayList<List<Integer>>();if (root == null) return levels;Queue<TreeNode> queue = new LinkedList<TreeNode>();queue.add(root);int level = 0;while ( !queue.isEmpty() ) {// start the current levellevels.add(new ArrayList<Integer>());// number of elements in the current levelint level_length = queue.size();for(int i = 0; i < level_length; ++i) {TreeNode node = queue.remove();// fulfill the current levellevels.get(level).add(node.val);// add child nodes of the current level// in the queue for the next levelif (node.left != null) queue.add(node.left);if (node.right != null) queue.add(node.right);}// go to next levellevel++;}return levels;}
}

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