102. Binary Tree Level Order Traversal.md

102. Binary Tree Level Order Traversal

Question:

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]
]

Thinking:

• Method:
  • 很明显的bfs,唯一需要注意的就是如何确定level

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if(root == null) return result;
        LinkedList<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        while(queue.size() != 0){
            int count = queue.size();
            List<Integer> list = new LinkedList<>();
            for(int i = 0; i < count; i++){
                TreeNode node = queue.poll();
                if(node.left != null) queue.add(node.left);
                if(node.right != null) queue.add(node.right);
                list.add(node.val);
            }
            result.add(list);
        }
        return result;
    }
}

二刷

这道题还是比较简单，使用bfs解就好了。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new LinkedList<>();
        if(root == null) return result;
        LinkedList<TreeNode> q = new LinkedList<>();
        q.offer(root);
        while(!q.isEmpty()){
            int size = q.size();
            List<Integer> res = new LinkedList<>();
            TreeNode node = null;
            for(int i = 0; i < size; i++){
                node = q.poll();
                res.add(node.val);
                if(node.left != null) q.offer(node.left);
                if(node.right != null) q.offer(node.right);
            }
            result.add(res);
        }
        return result;
    }
}

Third Time

• Method 1: bfs

  /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode(int x) { val = x; }
   * }
   */
  class Solution {
      public List<List<Integer>> levelOrder(TreeNode root) {
          List<List<Integer>> result = new ArrayList<>();
          if(root == null) return result;
          Queue<TreeNode> q = new LinkedList<>();
          q.offer(root);
          while(!q.isEmpty()){
              List<Integer> temp = new ArrayList<>();
              int size = q.size();
              for(int i = 0; i < size; i++){
                  TreeNode node = q.poll();
                  temp.add(node.val);
                  if(node.left != null) q.offer(node.left);
                  if(node.right != null) q.offer(node.right);
              }
              result.add(temp);
          }
          return result;
      }
  }

Fourth Time

• Method 1: BFS

   /**
    * Definition for a binary tree node.
    * public class TreeNode {
    *     int val;
    *     TreeNode left;
    *     TreeNode right;
    *     TreeNode(int x) { val = x; }
    * }
    */
   class Solution {
   	public List<List<Integer>> levelOrder(TreeNode root) {
   		List<List<Integer>> result = new ArrayList<>();
   		if(root == null) return result;
   		Queue<TreeNode> q = new LinkedList<>();
   		q.offer(root);
   		while(!q.isEmpty()){
   			List<Integer> temp = new ArrayList<>();
   			int size = q.size();
   			for(int i = 0; i < size; i++){
   				TreeNode node = q.poll();
   				temp.add(node.val);
   				if(node.left != null) q.offer(node.left);
   				if(node.right != null) q.offer(node.right);
   			}
   			result.add(temp);
   		}
   		return result;
   	}
   }