515. Find Largest Value in Each Tree Row
Given the root of a binary tree, return an array of the largest value in each row of the tree (0-indexed).
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| Example 1:
Input: root = [1,3,2,5,3,null,9]
Output: [1,3,9]
Example 2:
Input: root = [1,2,3]
Output: [1,3]
Example 3:
Input: root = [1]
Output: [1]
Example 4:
Input: root = [1,null,2]
Output: [1,2]
Example 5:
Input: root = []
Output: []
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Constraints:
- The number of nodes in the tree will be in the range [0, 104].
- -231 <= Node.val <= 231 - 1
Solution#
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| /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> largestValues(TreeNode root) {
Queue<TreeNode> q = new LinkedList<>();
if (root != null) {
q.add(root);
}
List<Integer> res = new ArrayList<>();
while (q.size() > 0) {
int size = q.size();
int max = Integer.MIN_VALUE;
for (int i = 0; i < size; i++) {
TreeNode node = q.poll();
max = Math.max(node.val, max);
if (node.left != null) q.add(node.left);
if (node.right != null) q.add(node.right);
}
res.add(max);
}
return res;
}
public List<Integer> largestValues2(TreeNode root) {
List<Integer> res = new ArrayList<>();
dfs(root, res, 0);
return res;
}
void dfs(TreeNode node, List<Integer> res, int level) {
if (node == null) return;
if (res.size() == level) {
res.add(node.val);
}
Integer max = Math.max(res.get(level), node.val);
res.set(level, max);
dfs(node.left, res, level + 1);
dfs(node.right, res, level + 1);
}
}
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Solution 2021-10-20#
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| class Solution {
public List<Integer> largestValues(TreeNode root) {
if (root == null) return Collections.emptyList();
Queue<TreeNode> q = new ArrayDeque<>();
List<Integer> res = new ArrayList<>();
q.add(root);
while (q.size() > 0) {
int size = q.size();
int max = Integer.MIN_VALUE;
for (int i = 0; i < size; i++) {
TreeNode node = q.poll();
max = Math.max(node.val, max);
if (node.left != null) {
q.add(node.left);
}
if (node.right != null) {
q.add(node.right);
}
}
res.add(max);
}
return res;
}
}
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