102. 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]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

链接： http://leetcode.com/problems/binary-tree-level-order-traversal/

题解：

层序遍历二叉树。使用BFS, 用一个Queue来辅助存储当前层的节点。

Time Complexity - O(n)， Space Complexity - O(n) (最多一层节点数)

public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(root == null)
            return result;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        ArrayList<Integer> list = new ArrayList<Integer>();
        queue.add(root);
        int curLevel = 1, nextLevel = 0;
        
        while(!queue.isEmpty()){
            TreeNode node = queue.poll();
            curLevel --;
            list.add(node.val);
            
            if(node.left != null){
                queue.offer(node.left);
                nextLevel ++;
            }
            if(node.right != null){
                queue.offer(node.right);
                nextLevel ++;
            }
            if(curLevel == 0){
                curLevel = nextLevel;
                nextLevel = 0;
                result.add(new ArrayList<Integer>(list));
                list.clear();
            }
        }
        
        return result;
    }
}

Update:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null)
            return res;
        Queue<TreeNode> queue = new LinkedList<>();         //BFS
        ArrayList<Integer> list = new ArrayList<>();
        queue.add(root);                                    //Initialize
        int curLevel = 1, nextLevel = 0;
        
        while(!queue.isEmpty()) {
            TreeNode node = queue.poll();
            list.add(node.val);
            curLevel --;
            if(node.left != null) {
                queue.offer(node.left);
                nextLevel++;
            }
            if(node.right != null) {
                queue.offer(node.right);
                nextLevel++;
            }
            if(curLevel == 0) {
                curLevel = nextLevel;
                nextLevel = 0;
                res.add(new ArrayList<Integer>(list));
                list.clear();
            }
        }
        
        return res;
    }
}

 二刷:

使用一个queue来帮助BFS层序遍历，有两个记录当前层节点和下一层节点数的变量curLevel以及nextLevel。

Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int curLevel = 1, nextLevel = 0;
        List<Integer> list = new ArrayList<>();
        while (!q.isEmpty()) {
            TreeNode node = q.poll();
            curLevel--;
            list.add(node.val);
            if (node.left != null) {
                q.offer(node.left);
                nextLevel++;
            }
            if (node.right != null) {
                q.offer(node.right);
                nextLevel++;
            }
            if (curLevel == 0) {
                curLevel = nextLevel;
                nextLevel = 0;
                res.add(new ArrayList<Integer>(list));
                list.clear();
            }
        }
        return res;
    }
}

三刷:

不要忘记把结果保存到结果集里，并且也要clear当前层。

Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        List<Integer> level = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int curLevel = 1, nextLevel = 0;
        
        while (!q.isEmpty()) {
            TreeNode node = q.poll();
            curLevel--;
            level.add(node.val);
            if (node.left != null) {
                q.offer(node.left);
                nextLevel++;
            }
            if (node.right != null) {
                q.offer(node.right);
                nextLevel++;
            }
            if (curLevel == 0) {
                curLevel = nextLevel;
                nextLevel = 0;
                res.add(new ArrayList<Integer>(level));
                level.clear();
            }
        }
        return res;
    }
}

Update: 换种写法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) return res;
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int curLevel = 1;
        List<Integer> list = new ArrayList<>();
        
        while (!q.isEmpty()) {
            TreeNode node = q.poll();
            curLevel--;
            list.add(node.val);
            if (node.left != null) q.offer(node.left);
            if (node.right != null) q.offer(node.right);
            if (curLevel == 0) {
                curLevel = q.size();
                res.add(new ArrayList<>(list));
                list.clear();
            }    
        }
        
        return res;
    }
}

Reference:

https://leetcode.com/discuss/21778/java-solution-using-dfs

https://leetcode.com/discuss/22533/java-solution-with-a-queue-used

https://leetcode.com/discuss/58454/java-queue-solution-beats-100%25