Binary Tree Level Order Traversal II

ID: 70; medium

LintCode 炼码

Solution 1 (Java)

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param root: A tree
     * @return: buttom-up level order a list of lists of integer
     */
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> ans = new ArrayList<>();
        // List<List<Integer>> temp = new ArrayList<>();
        if (root == null) return ans;
        Queue<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        
        while (!q.isEmpty()) {
            int n = q.size();
            List<Integer> level = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                TreeNode curr = q.poll();
                level.add(curr.val);

                if (curr.left != null) q.offer(curr.left);
                if (curr.right != null) q.offer(curr.right);
            }
            ans.add(0, level);
        }
        
        return ans;
    }
}