有点感动,这题在AS里写好之后复制过去一次AC了。
这种BFS的套路跟这题的第一个版本一样,用queue,维护curNum和nextNum;我跟code ganker学会了这个套路,按照覃超的标准来说这代码是有点长的,但是现在已经形成记忆了,就不去修改了。思路还蛮清晰的。
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) return res;
LinkedList<TreeNode> queue = new LinkedList<>();
List<Integer> cell = new ArrayList<>();
queue.add(root);
int curNum = 1;
int nextNum = 0;
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
cell.add(node.val);
curNum--;
if (node.left != null) {
queue.offer(node.left);
nextNum++;
}
if (node.right != null) {
queue.offer(node.right);
nextNum++;
}
if (curNum == 0) {
res.add(0, cell);
curNum = nextNum;
nextNum = 0;
cell = new ArrayList<>();
}
}
return res;
}
另
贴一下leetcode里面的人的简洁写法:https://discuss.leetcode.com/topic/7651/my-dfs-and-bfs-java-solution/2
public class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
Queue<TreeNode> queue = new LinkedList<TreeNode>();
List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
if(root == null) return wrapList;
queue.offer(root);
while(!queue.isEmpty()){
int levelNum = queue.size();
List<Integer> subList = new LinkedList<Integer>();
for(int i=0; i<levelNum; i++) {
if(queue.peek().left != null) queue.offer(queue.peek().left);
if(queue.peek().right != null) queue.offer(queue.peek().right);
subList.add(queue.poll().val);
}
wrapList.add(0, subList);
}
return wrapList;
}
}
递归(我没看):
public class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
levelMaker(wrapList, root, 0);
return wrapList;
}
public void levelMaker(List<List<Integer>> list, TreeNode root, int level) {
if(root == null) return;
if(level >= list.size()) {
list.add(0, new LinkedList<Integer>());
}
levelMaker(list, root.left, level+1);
levelMaker(list, root.right, level+1);
list.get(list.size()-level-1).add(root.val);
}
}
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