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  • import java.util.ArrayList;
    import java.util.LinkedList;
    import java.util.List;
    import java.util.Queue;
    
    /**
    
     You need to find the largest value in each row of a binary tree.
    
     Example:
     Input:
    
        1
       / \
      3   2
     / \   \
    5   3   9
    
     Output: [1, 3, 9]
    
     */
    
    public class Find_Largest_Value_in_Each_Tree_Row {
    
        /**
         * Definition for a binary tree node.
         * public class TreeNode {
         *     int val;
         *     TreeNode left;
         *     TreeNode right;
         *     TreeNode(int x) { val = x; }
         * }
         */
        class Solution {
            public List<Integer> largestValues(TreeNode root) {
    
                List<Integer> result = new ArrayList<>();
    
                // bfs
                if (root == null) {
                    return result;
                }
    
                Queue<TreeNode> q = new LinkedList<>();
                q.offer(root);
                int currentLevelCount = 1;
                int nextLevelCount = 0;
    
                int currentLevelMax = root.val;
    
                while (!q.isEmpty()) {
                    TreeNode current = q.poll();
                    currentLevelCount--;
    
                    // check leftmost
                    currentLevelMax = Math.max(currentLevelMax, current.val);
    
                    if (current.left != null) {
                        q.offer(current.left);
                        nextLevelCount++;
                    }
                    if (current.right != null) {
                        q.offer(current.right);
                        nextLevelCount++;
                    }
    
                    if (currentLevelCount == 0) {
                        currentLevelCount = nextLevelCount;
                        nextLevelCount = 0;
    
                        result.add(currentLevelMax);
    
                        // reset
                        currentLevelMax = Integer.MIN_VALUE;
                    }
                }
    
                return result;
            }
        }
    }
    
    ############
    
    /**
     * 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) {
            List<Integer> ans = new ArrayList<>();
            if (root == null) {
                return ans;
            }
            Deque<TreeNode> q = new ArrayDeque<>();
            q.offer(root);
            while (!q.isEmpty()) {
                int t = q.peek().val;
                for (int i = q.size(); i > 0; --i) {
                    TreeNode node = q.poll();
                    t = Math.max(t, node.val);
                    if (node.left != null) {
                        q.offer(node.left);
                    }
                    if (node.right != null) {
                        q.offer(node.right);
                    }
                }
                ans.add(t);
            }
            return ans;
        }
    }
    
  • // OJ: https://leetcode.com/problems/find-largest-value-in-each-tree-row/
    // Time: O(N)
    // Space: O(N)
    class Solution {
    public:
        vector<int> largestValues(TreeNode* root) {
            if (!root) return {};
            vector<int> ans;
            queue<TreeNode*> q{ {root} };
            while (q.size()) {
                int cnt = q.size(), maxVal = INT_MIN;
                while (cnt--) {
                    auto n = q.front();
                    q.pop();
                    maxVal = max(maxVal, n->val);
                    if (n->left) q.push(n->left);
                    if (n->right) q.push(n->right);
                }
                ans.push_back(maxVal);
            }
            return ans;
        }
    };
    
  • # Definition for a binary tree node.
    # class TreeNode:
    #     def __init__(self, val=0, left=None, right=None):
    #         self.val = val
    #         self.left = left
    #         self.right = right
    class Solution:
        def largestValues(self, root: Optional[TreeNode]) -> List[int]:
            if root is None:
                return []
            q = deque([root])
            ans = []
            while q:
                t = -inf
                for _ in range(len(q)):
                    node = q.popleft()
                    t = max(t, node.val)
                    if node.left:
                        q.append(node.left)
                    if node.right:
                        q.append(node.right)
                ans.append(t)
            return ans
    
    ############
    
    # Definition for a binary tree node.
    # class TreeNode(object):
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    
    class Solution(object):
      def largestValues(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        ans = []
        d = {}
    
        def dfs(root, h, d):
          if root:
            dfs(root.left, h + 1, d)
            dfs(root.right, h + 1, d)
            d[h] = max(d.get(h, float("-inf")), root.val)
    
        dfs(root, 0, d)
        level = 0
        while level in d:
          ans += d[level],
          level += 1
        return ans
    
    
  • /**
     * Definition for a binary tree node.
     * type TreeNode struct {
     *     Val int
     *     Left *TreeNode
     *     Right *TreeNode
     * }
     */
    func largestValues(root *TreeNode) []int {
    	var ans []int
    	if root == nil {
    		return ans
    	}
    	q := []*TreeNode{root}
    	for len(q) > 0 {
    		t := q[0].Val
    		for i := len(q); i > 0; i-- {
    			node := q[0]
    			q = q[1:]
    			t = max(t, node.Val)
    			if node.Left != nil {
    				q = append(q, node.Left)
    			}
    			if node.Right != nil {
    				q = append(q, node.Right)
    			}
    		}
    		ans = append(ans, t)
    	}
    	return ans
    }
    
    func max(a, b int) int {
    	if a > b {
    		return a
    	}
    	return b
    }
    
  • /**
     * Definition for a binary tree node.
     * class TreeNode {
     *     val: number
     *     left: TreeNode | null
     *     right: TreeNode | null
     *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
     *         this.val = (val===undefined ? 0 : val)
     *         this.left = (left===undefined ? null : left)
     *         this.right = (right===undefined ? null : right)
     *     }
     * }
     */
    
    function largestValues(root: TreeNode | null): number[] {
        const res: number[] = [];
        const queue: TreeNode[] = [];
        if (root) {
            queue.push(root);
        }
        while (queue.length) {
            const n = queue.length;
            let max = -Infinity;
            for (let i = 0; i < n; i++) {
                const { val, left, right } = queue.shift();
                max = Math.max(max, val);
                left && queue.push(left);
                right && queue.push(right);
            }
            res.push(max);
        }
        return res;
    }
    
    
  • // Definition for a binary tree node.
    // #[derive(Debug, PartialEq, Eq)]
    // pub struct TreeNode {
    //   pub val: i32,
    //   pub left: Option<Rc<RefCell<TreeNode>>>,
    //   pub right: Option<Rc<RefCell<TreeNode>>>,
    // }
    //
    // impl TreeNode {
    //   #[inline]
    //   pub fn new(val: i32) -> Self {
    //     TreeNode {
    //       val,
    //       left: None,
    //       right: None
    //     }
    //   }
    // }
    use std::rc::Rc;
    use std::cell::RefCell;
    use std::collections::VecDeque;
    impl Solution {
        pub fn largest_values(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
            let mut res = Vec::new();
            let mut queue = VecDeque::new();
            if root.is_some() {
                queue.push_back(root.clone());
            }
            while !queue.is_empty() {
                let mut max = i32::MIN;
                for _ in 0..queue.len() {
                    let node = queue.pop_front().unwrap();
                    let node = node.as_ref().unwrap().borrow();
                    max = max.max(node.val);
                    if node.left.is_some() {
                        queue.push_back(node.left.clone());
                    }
                    if node.right.is_some() {
                        queue.push_back(node.right.clone());
                    }
                }
                res.push(max);
            }
            res
        }
    }
    
    

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