LeetCode | Binary Tree Zigzag Level Order Traversal

题目

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example:

Given binary tree 

{3,9,20,#,#,15,7}

,

3
/ \
9  20
/  \
15   7

return its zigzag level order traversal as:

[
[3],
[20,9],
[15,7]
]

分析
从遍历顺序上看，直接想到BFS和栈，实现时遍历顺序还是需要耐心确认的。

代码

import java.util.ArrayList;
import java.util.Stack;

public class BinaryTreeZigzagLevelOrderTraversal {
public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
if (root == null) {
return results;
}
Stack<TreeNode> currentLevel = new Stack<TreeNode>();
Stack<TreeNode> nextLevel = new Stack<TreeNode>();
boolean reverse = false;
ArrayList<Integer> list = new ArrayList<Integer>();
currentLevel.push(root);
while (!currentLevel.isEmpty()) {
TreeNode node = currentLevel.pop();
list.add(node.val);
if (reverse) {
if (node.right != null) {
nextLevel.push(node.right);
}
if (node.left != null) {
nextLevel.push(node.left);
}
} else {
if (node.left != null) {
nextLevel.push(node.left);
}
if (node.right != null) {
nextLevel.push(node.right);
}
}

if (currentLevel.isEmpty()) {
Stack<TreeNode> temp = currentLevel;
currentLevel = nextLevel;
nextLevel = temp;
results.add(new ArrayList<Integer>(list));
list.clear();
reverse = !reverse;
}
}
return results;
}
}