Leetcode102.二叉树的层次遍历
非递归队列做法:
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
ArrayList<List<Integer>> result = new ArrayList();
if(root == null){
return result;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
int level = 0;
while ( !queue.isEmpty() ) {
result.add(new ArrayList<Integer>());
int level_length = queue.size();
for(int i = 0; i < level_length; i++) {
TreeNode node = queue.remove();
result.get(level).add(node.val);
if (node.left != null) queue.add(node.left);
if (node.right != null) queue.add(node.right);
}
level++;
}
return result;
}
} 递归: ArrayList<List<Integer>> result = new ArrayList();
public void func(TreeNode node, int level) {
if (result.size() == level)
result.add(new ArrayList<Integer>());
result.get(level).add(node.val);
if (node.left != null)
func(node.left, level + 1);
if (node.right != null)
func(node.right, level + 1);
}
public List<List<Integer>> levelOrder(TreeNode root) {
if (root == null) {
return result;
}
func(root, 0);
return result;
} PS:递归的内存消耗居然要低于非递归,不知是不是测试样例的问题。。
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