103. Binary Tree Zigzag Level Order Traversal

103. Binary Tree Zigzag Level Order Traversal

Solution 1: public class Solution {     public List<List<Integer>> zigzagLevelOrder(TreeNode root) {         List<List<Integer>> sol = new ArrayList<>();         travel(root, sol, 0);         return sol;     }          private void travel(TreeNode curr, List<List<Integer>> sol, int level) {         if (curr == null)             return;         if (sol.size() <= level) {                         sol.add(new ArrayList<>());         }                  if (level % 2 == 0)             sol.get(level).add(curr.val);         else              sol.get(level).add(0, curr.val); //在0位置直接add,所以原来的元素右移                  travel(curr.left, sol, level + 1);         travel(curr.right, sol, level + 1);     } } Solution 2: public class Solution {   public List<List<Integer>> zigzagLevelOrder(TreeNode root) {     List<List<Integer>> res = new ArrayList<>();     if (root == null) return res;       Queue<TreeNode> q = new LinkedList<>();     q.add(root);     boolean order = true;       while(!q.isEmpty()) {         int size = q.size();         List<Integer> tmp = new ArrayList<>();         for(int i = 0; i < size; ++i) {             TreeNode n = q.poll();             if(order) {                 tmp.add(n.val);             } else {                 tmp.add(0, n.val);             }             if(n.left != null) q.add(n.left);             if(n.right != null) q.add(n.right);         }         res.add(tmp);         order = order ? false : true;     }     return res;  } }

 

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