102 Binary Tree Level Order Traversal – Easy
Problem:
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]
]
Thoughts:
This is a bread first search order traversal.
Use two queues to remember which level currently at.
Solutions:
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
Queue<TreeNode> qv = new LinkedList<TreeNode>();
List<List<Integer>> result = new LinkedList<List<Integer>>();
if (root == null) {
return result;
}
qv.add(root);
while (qv.size() > 0) {
int n = qv.size();
List<Integer> add = new LinkedList<Integer>();
for (int i = 0; i < n; i ++) {
TreeNode node = qv.remove();
add.add(node.val);
if (node.left !=null) {
qv.add(node.left);
}
if (node.right !=null) {
qv.add(node.right);
}
}
result.add(add);
}
return result;
}//list
}