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] ]
confused what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.
Analysis: It is similar as the binary tree level order traversal, i.e., use BFS. However, the frontier and next here are more suitable for using Stack instead of Queue. We just use a boolean variable as the flag to determine the order to put into the stack (left to right, or, say, right to left).
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
if(root==null) return res;
boolean leftToRight = false;
Stack<TreeNode> frontier = new Stack<TreeNode>();
frontier.push(root);
while(!frontier.isEmpty()) {
Stack<TreeNode> next = new Stack<TreeNode>();
ArrayList<Integer> tem = new ArrayList<Integer>();
while(!frontier.isEmpty()) {
TreeNode current = frontier.pop();
tem.add(current.val);
if(!leftToRight) { // right to left traversal
if(current.left!=null) next.push(current.left);
if(current.right!=null) next.push(current.right);
}
else { // left to right traversal
if(current.right!=null) next.push(current.right);
if(current.left!=null) next.push(current.left);
}
}
res.add(tem);
leftToRight = !leftToRight;
frontier = next;
}
return res;
}
}