LeetCode 107. Binary Tree Level Order Traversal II

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        return bfs(root);
    }
    
    private List<List<Integer> > bfs(TreeNode root) {
        List<List<Integer> > ans = new ArrayList<>();
        if (root == null) return ans;
        
        Queue<TreeNode> Q = new LinkedList<TreeNode>();
        Q.add(root);
        
        TreeNode endRight = root;     //记录当前深度最右边的节点
        TreeNode currentRight = null; //记录当前遍历的节点中最右边的孩子节点
        
        List<Integer> list = new ArrayList<>();
        
        while (!Q.isEmpty()) {
            TreeNode t = Q.poll();
            list.add(t.val);
            
            if(t.left != null) {
                Q.add(t.left);
                currentRight = t.left;
            }
            if(t.right != null) {
                Q.add(t.right);
                currentRight = t.right;
            }
            if(t == endRight) {   //如果当前节点等于当前深度最右边的节点，那么深度就要＋1了
                endRight = currentRight;
                ans.add(0, list);
                list = new ArrayList<>();
            }
        }
        return ans;
    }
}