leetcode 107. Binary Tree Level Order Traversal II

107. Binary Tree Level Order Traversal II:题目链接

BFS

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    // 层序遍历这棵树  将第i层的结点全部加入到一个list中  最后再将该list压入栈中  最后将栈中结果集出栈就是answer
    // 其实也可以不用栈  用java里面的一个方法 res.add(0, 该层的结点集合) 就可以起到压栈再出栈的效果
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res= new ArrayList<>();
        if (root == null) {
            return res;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        Stack<List<Integer>> stack = new Stack<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            List<Integer> list = new ArrayList<>();
            for (int size = queue.size(); size > 0; --size) {
                TreeNode front = queue.poll();
                list.add(front.val);
                if (front.left != null) {
                    queue.offer(front.left);
                }
                if (front.right != null) {
                    queue.offer(front.right);
                }
            }
            stack.push(list);
        }
        
        while (!stack.isEmpty()) {
            res.add(stack.pop());
        }
        return res;
    }
}