Wednesday, February 12, 2014

Binary Tree Level Order Traversal @LeetCode

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]
]

package leetcode.tree;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
/**
* Thought: Use Queue to store every layer nodes, the most important is to terminate level. we use queue size.
*
* 1. int queueSize = queue.size(), we cannot put queue.size() in for loop as it will changes in loop.
* 2. Queue is just a interface, we use LinkedList as implement class.
* 3. Must judge if root == null, or NullPointerProblem in list.add(node.val). return result instead of null -- coding style.
* Extension: Except this way, any other ways?
* 1. Queue + dummyNode --> insert dummyNode when one line ends. dummyNode has No meaning like Integer.MAX_VALUE
* 2. Two Queue --> next time, we stored in another queue, for loop ends until size of current queue size==0.
*
* Actually, our ways is the best way (use quese size to control line).
*
* @author jeffwan
* @date Feb 12, 2014
*/
public class LevelOrder {
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>();
queue.offer(root);
while(!queue.isEmpty()) {
ArrayList<Integer> level = new ArrayList<Integer>();
int queueSize = queue.size();
for (int i = 0 ; i < queueSize; i++) {
TreeNode node = queue.poll();
level.add(node.val);
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
result.add(level);
}
return result;
}
// TreeNode
private class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode (int x) {
val = x;
}
}
}
view raw LevelOrder.java hosted with ❤ by GitHub

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