采用 BFS 算法逐行遍历，JAVA 解法：

    public final List<Integer> largestValues(TreeNode root) {
        List<Integer> reList = new LinkedList<Integer>();
        if (root == null) return reList;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        while (!queue.isEmpty()) {
            int currentSize = queue.size();
            int currentMax = Integer.MIN_VALUE;
            for (int i = 0; i < currentSize; i++) {
                TreeNode node = queue.poll();
                currentMax = Math.max(currentMax, node.val);
                if (node.left != null) queue.add(node.left);
                if (node.right != null) queue.add(node.right);
            }
            reList.add(currentMax);
        }
        return reList;
    }

　　JS 解法：

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var largestValues = function (root) {
    if (!root) return [];
    let reList = [];
    let nodeList = [];
    nodeList.push(root);
    while (nodeList.length > 0) {
        let currentMax = -Number.MAX_VALUE;
        let currentSize = nodeList.length;
        for (let i = 0; i < currentSize; i++) {
            let node = nodeList.shift();
            currentMax = currentMax > node.val ? currentMax : node.val;
            if (node.left) nodeList.push(node.left);
            if (node.right) nodeList.push(node.right);
        }
        reList.push(currentMax);
    }
    return reList;
};