103. Binary Tree Zigzag Level Order Traversal

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
      List<List<Integer>> result = new ArrayList<>();
      if(root == null){
        return result;
      }
      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()){
          TreeNode current = s1.pop();
          tmp.add(current.val);
          if(current.left != null){
            s2.push(current.left);
          }
          if(current.right != null){
            s2.push(current.right);
          }
          
        }
        result.add(tmp);
        List<Integer> tmp2 = new ArrayList<>();
        while(!s2.isEmpty()){
          TreeNode current = s2.pop();
          tmp2.add(current.val);
          if(current.right != null){
            s1.push(current.right);
          }
          if(current.left != null){
            s1.push(current.left);
          }
          
        }
        if(!tmp2.isEmpty()){
          result.add(tmp2);
        }
      }
      return result;
    }
}

 

 

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]
]