LeetCode(107)——Binary Tree Level Order Traversal II

題目：

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]
]

和102類似，只不過每層添加到最後列表的順序反過來了。

 

AC:

public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> resultList = new LinkedList<>();
        
        if (root == null) {
            return resultList;
        }
        
        List<TreeNode> tmp = new LinkedList<>();
        tmp.add(root);
        
        while (!tmp.isEmpty()) {
            List<Integer> valueList = new ArrayList<>();
            int size = tmp.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = tmp.get(0);
                
                if (node.left != null) {
                    tmp.add(node.left);
                }
                if (node.right != null) {
                    tmp.add(node.right);
                }
                
                valueList.add(node.val);
                tmp.remove(0);
            }
            
            resultList.add(0, valueList);
        }
        
        return resultList;
    }

迭代，4ms

public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> resultList = new LinkedList<>();
        
        helper(resultList, root, 1);
        
        return resultList;
    }
    
    private void helper(List<List<Integer>> resultList, TreeNode root, int depth) {
        if (root == null) {
            return;
        }

        if (resultList.size() < depth) {
            resultList.add(0, new ArrayList<>());
        }
        
        resultList.get(resultList.size() - depth).add(root.val);
        
        helper(resultList, root.left, depth + 1);
        
        helper(resultList, root.right, depth + 1);
    }

遞歸，3ms