[LeetCode]--102. Binary Tree Level Order Traversal

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] ] public List<List<Integer>> levelOrder(TreeNode root) { List<Integer> list1 = new ArrayList<Integer>(); List<List<Integer>> list = new ArrayList<List<Integer>>(); if (root == null) return list; Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.add(root); int sum = 1, i = 1; while (!queue.isEmpty()) { TreeNode t1 = queue.poll(); System.out.println(t1.val); list1.add(t1.val); if (sum == Math.pow(2, i) - 1) { list.add(list1); list1 = new ArrayList<Integer>(); i++; } if (t1.left != null) queue.add(t1.left); if (t1.right != null) queue.add(t1.right); sum += 2; } if (!list1.isEmpty()) { list.add(list1); } return list; }

这个是我最开始的思路，我想用每一层sum计数，然后sum得到2的层次方减一就是在那一层这样的方法来控制每次list1要装几次。第一个错误，就是List1每次必须是new出来，用clear不行的；第二个错误，就是如果一个父节点压根就没有右节点或者没有左节点，那么上面的判断方式失效。比如：[3,9,20,null,null,15,7]。

后来我考虑到，每次加到队列中的元素其实就是那一层的元素，用queue的size()方法就行。这个要注意我标识在代码中了。

public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> list = new ArrayList<List<Integer>>(); if (root == null) return list; Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.add(root); while (!queue.isEmpty()) { List<Integer> list1 = new ArrayList<Integer>(); //这个注意，一定要先记录下来，不然底下size会变的 int size = queue.size(); for (int i = 0; i < size; i++) { TreeNode t1 = queue.poll(); list1.add(t1.val); if (t1.left != null) queue.add(t1.left); if (t1.right != null) queue.add(t1.right); } list.add(list1); } return list; }

这里我又去网上找了几种算法，第一种跟我基本一样。其余的大家可以多作了解，我有时间再看，先贴上来。

// version 1: BFS public class Solution { public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) { ArrayList result = new ArrayList(); if (root == null) { return result; } Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.offer(root); while (!queue.isEmpty()) { ArrayList<Integer> level = new ArrayList<Integer>(); int size = queue.size(); for (int i = 0; i < size; i++) { TreeNode head = queue.poll(); level.add(head.val); if (head.left != null) { queue.offer(head.left); } if (head.right != null) { queue.offer(head.right); } } result.add(level); } return result; } } // version 2: DFS public class Solution { /** * @param root: The root of binary tree. * @return: Level order a list of lists of integer */ public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) { ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>(); if (root == null) { return results; } int maxLevel = 0; while (true) { ArrayList<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, ArrayList<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); } } // version 3: BFS. two queues public class Solution { /** * @param root: The root of binary tree. * @return: Level order a list of lists of integer */ public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) { ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); if (root == null) { return result; } ArrayList<TreeNode> Q1 = new ArrayList<TreeNode>(); ArrayList<TreeNode> Q2 = new ArrayList<TreeNode>(); Q1.add(root); while (Q1.size() != 0) { ArrayList<Integer> level = new ArrayList<Integer>(); Q2.clear(); for (int i = 0; i < Q1.size(); i++) { TreeNode node = Q1.get(i); level.add(node.val); if (node.left != null) { Q2.add(node.left); } if (node.right != null) { Q2.add(node.right); } } // swap q1 and q2 ArrayList<TreeNode> temp = Q1; Q1 = Q2; Q2 = temp; // add to result result.add(level); } return result; } } // version 4: BFS, queue with dummy node public class Solution { /** * @param root: The root of binary tree. * @return: Level order a list of lists of integer */ public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) { ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); if (root == null) { return result; } Queue<TreeNode> Q = new LinkedList<TreeNode>(); Q.offer(root); Q.offer(null); // dummy node ArrayList<Integer> level = new ArrayList<Integer>(); while (!Q.isEmpty()) { TreeNode node = Q.poll(); if (node == null) { if (level.size() == 0) { break; } result.add(level); level = new ArrayList<Integer>(); Q.offer(null); // add a new dummy node continue; } level.add(node.val); if (node.left != null) { Q.offer(node.left); } if (node.right != null) { Q.offer(node.right); } } return result; } }