You need to find the largest value in each row of a binary tree.
Solution1：DFS
思路：
Time Complexity: O(N) Space Complexity: O(N)递归缓存
Solution2：BFS
思路：
Time Complexity: O(N) Space Complexity: O(N)
Solution1 Code：
public class Solution {
public List<Integer> largestValues(TreeNode root) {
List<Integer> result = new ArrayList<Integer>();
dfs(root, result, 0);
return result;
}
private void dfs(TreeNode root, List<Integer> result, int d){
if(root == null) {
return;
}
// expand list size
if(d == result.size()){
result.add(root.val);
}
else{
// or set value
result.set(d, Math.max(result.get(d), root.val));
}
dfs(root.left, result, d + 1);
dfs(root.right, result, d + 1);
}
}
Solution2 Code：
class Solution {
public List<Integer> largestValues(TreeNode root) {
List<Integer> result = new ArrayList<Integer>();
if(root == null) return result;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int queue_size = queue.size();
int max_value = Integer.MIN_VALUE;
for (int i = 0; i < queue_size; i++) {
TreeNode cur = queue.poll();
max_value = Math.max(cur.val, max_value);
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
result.add(max_value);
}
return result;
}
}
