[leetcode]103. Binary Tree Zigzag Level Order Traversal@Java解题报告

https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/description/

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,null,null,15,7]

,

3
/ \
9  20
/  \
15   7

return its zigzag level order traversal as:

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

package go.jacob.day806;

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Stack;

/**
*
* @author Jacob
* 103. Binary Tree Zigzag Level Order Traversal
*
* 两种解法：1.使用两个堆栈；2.使用一个queue，用flag记录当前行打印顺序
*/
public class Demo2 {
/*
* 方法一 解法与Binary Tree Level Order Traversal一致 区别是需要用一个flag判断应该正序or逆序打印
*/
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if (root == null)
return res;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
// 用flag控制正序或者逆序打印
boolean flag = true;
while (!queue.isEmpty()) {
int size = queue.size();
ArrayList<Integer> list = new ArrayList<Integer>();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
if (flag)
list.add(node.val);
else
list.add(0, node.val);

if (node.left != null)
queue.offer(node.left);
if (node.right != null)
queue.offer(node.right);
}
flag = !flag;
res.add(list);
}
return res;
}

/*
* 解法二 使用两个堆栈
*/
public List<List<Integer>> zigzagLevelOrder_2(TreeNode root) {
TreeNode c = root;
List<List<Integer>> ans = new ArrayList<List<Integer>>();
if (c == null)
return ans;
Stack<TreeNode> s1 = new Stack<TreeNode>();
Stack<TreeNode> s2 = new Stack<TreeNode>();
s1.push(root);
while (!s1.isEmpty() || !s2.isEmpty()) {
List<Integer> tmp = new ArrayList<Integer>();
while (!s1.isEmpty()) {
c = s1.pop();
tmp.add(c.val);
if (c.left != null)
s2.push(c.left);
if (c.right != null)
s2.push(c.right);
}
ans.add(tmp);
tmp = new ArrayList<Integer>();
while (!s2.isEmpty()) {
c = s2.pop();
tmp.add(c.val);
if (c.right != null)
s1.push(c.right);
if (c.left != null)
s1.push(c.left);
}
if (!tmp.isEmpty())
ans.add(tmp);
}
return ans;
}

private class TreeNode {
int val;
TreeNode left;
TreeNode right;

TreeNode(int x) {
val = x;
}
}
}