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  • 107. Binary Tree Level Order Traversal II

    Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

    For example:
    Given binary tree [3,9,20,null,null,15,7],

        3
       / 
      9  20
        /  
       15   7

    return its bottom-up level order traversal as:

    [
      [15,7],
      [9,20],
      [3]
    ]

    题目含义:这道题的要求是从下往上分层遍历二叉树。
     1     public List<List<Integer>> levelOrderBottom(TreeNode root) {
     2         List<List<Integer>> result = new LinkedList<List<Integer>>();
     3         Queue<TreeNode> queue = new LinkedList<>();
     4         if (root == null) return result;
     5         queue.add(root);
     6         while (!queue.isEmpty()) {
     7             int size = queue.size();
     8             List<Integer> values = new ArrayList<>();
     9             for (int i = 0; i < size; i++) {
    10                 TreeNode node = queue.poll();
    11                 if (node.left != null) queue.offer(node.left);
    12                 if (node.right != null) queue.offer(node.right);
    13                 values.add(node.val);
    14             }
    15             result.add(0, values);
    16         }
    17         return result;     
    18     }
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  • 原文地址:https://www.cnblogs.com/wzj4858/p/7705373.html
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