标签：== ntc www traversal node ret str you algo
Given a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).
Example
Given binary tree {3,9,20,#,#,15,7},
3
/ 9 20
/ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
Challenge 1: Using only 1 queue to implement it.
Challenge 2: Use DFS algorithm to do it.
1. BFS：和queue组合做，while(q.notEmpty)每层，嵌套for(层里每个)做。每层的任务就是取出此层节点，加入层的list，并且把孩子们推进queue。
2.DFS：一层一层的list造起来。每次规定加maxLevel这层的节点。DFS到这一层的时候加节点；null或者超的时候不加；还没到这层就继续向下走。
1. BFS：
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /* * @param root: A Tree * @return: Level order a list of lists of integer */ public List<List<Integer>> levelOrder(TreeNode root) { // write your code here List<List<Integer>> result = new ArrayList<List<Integer>>(); if (root == null) { return result; } Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.offer(root); while (!queue.isEmpty()) { int size = queue.size(); List<Integer> crtLevel = new ArrayList<Integer>(); for (int i = 0; i < size; i++) { TreeNode node= queue.poll(); crtLevel.add(node.val); if (node.left != null) { queue.offer(node.left); } if (node.right != null) { queue.offer(node.right); } } result.add(crtLevel); } return result; } }
2.DFS
// version 2: DFS public class Solution { /** * @param root: The root of binary tree. * @return: Level order a list of lists of integer */ public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> results = new ArrayList<List<Integer>>(); if (root == null) { return results; } int maxLevel = 0; while (true) { List<Integer> level = new ArrayList<Integer>(); dfs(root, level, 0, maxLevel); if (level.size() == 0) { break; } results.add(level); maxLevel++; } return results; } private void dfs(TreeNode root, List<Integer> level, int curtLevel, int maxLevel) { if (root == null || curtLevel > maxLevel) { return; } if (curtLevel == maxLevel) { level.add(root.val); return; } dfs(root.left, level, curtLevel + 1, maxLevel); dfs(root.right, level, curtLevel + 1, maxLevel); } }
lintcode69- Binary Tree Level Order Traversal- easy
标签：== ntc www traversal node ret str you algo
原文地址：http://www.cnblogs.com/jasminemzy/p/7707726.html
