Binary Tree Level Order Traversal - leetcode - java

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]
]

DFS实现：

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
// DFS 遍历
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
traverse(res, root, 0);
return res;
}

public void traverse(List<List<Integer>> res, TreeNode root, int level) {
if (root == null) {
return;
}
if (res.size() > level) {
res.get(level).add(root.val);
} else {
ArrayList list = new ArrayList<Integer>();
list.add(root.val);
res.add(list);
}
traverse(res, root.left, level + 1);
traverse(res, root.right, level + 1);
}

}

BFS实现:

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
// BFS遍历打印
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new LinkedList<List<Integer>>();
if (root == null)
return res;
LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
ArrayList<Integer> list = new ArrayList<Integer>();
int cur = 1;
int nextLevel = 0;
while (!queue.isEmpty()) {
TreeNode tmp = queue.poll();

cur--;
if (tmp.left != null) {
queue.add(tmp.left);
nextLevel++;
}
if (tmp.right != null) {
queue.add(tmp.right);
nextLevel++;
}
list.add(tmp.val);
if (cur == 0) {
res.add(list);
list = new ArrayList<Integer>();
cur = nextLevel;
nextLevel = 0;
}

}
return res;
}

}