【Leetcode】Binary Tree Level Order Traversal

题目链接：https://leetcode.com/problems/binary-tree-level-order-traversal/

题目：

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,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its level order traversal as:

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

思路：

1、利用上文提到的 获取每层结点的方法 ，形似 Binary Tree Level Order Traversal II

2、利用层次遍历

算法1：

public List<List<Integer>> levelOrder(TreeNode root) {int height = heightTree(root);List<List<Integer>> lists = new ArrayList<List<Integer>>();for (int i = 1; i <= height; i++) {List<Integer> list = new ArrayList<Integer>();list = kLevelNumber(root, 1, list, i);lists.add(list);}return lists;}/** * 求第kk层节点 */public List<Integer> kLevelNumber(TreeNode p, int height, List<Integer> list, int kk) {if (p != null) {if (height == kk) {list.add(p.val);}list = kLevelNumber(p.left, height + 1, list, kk);list = kLevelNumber(p.right, height + 1, list, kk);}return list;}public int heightTree(TreeNode p) {if (p == null)return 0;int h1 = heightTree(p.left);int h2 = heightTree(p.right);return h1 > h2 ? h1 + 1 : h2 + 1;}

算法2：

public List<List<Integer>> levelOrder(TreeNode root) {List<List<Integer>> lists = new ArrayList<List<Integer>>();Queue<TreeNode> q = new LinkedList<TreeNode>();if (root == null)return lists;q.offer(root);while (!q.isEmpty()) {List<Integer> list = new ArrayList<Integer>();int size = q.size();for (int i = 0; i < size; i++) { // 该层结点数TreeNode t = q.poll();list.add(t.val);if (t.left != null) {q.offer(t.left);}if (t.right != null) {q.offer(t.right);}}lists.add(list);}return lists;}