在每个树行中找最大值

• https://leetcode.cn/problems/find-largest-value-in-each-tree-row/description/

描述

• 给定一棵二叉树的根节点 root ，请找出该二叉树中每一层的最大值

示例1

输入: root = [1,3,2,5,3,null,9]
输出: [1,3,9]

示例2

输入: root = [1,2,3]
输出: [1,3]

提示

• 二叉树的节点个数的范围是 [0,$10^4$]
• $-2^{31}$ <= Node.val <= $2^{31}$ - 1

Typescript 版算法实现

1 ) 方案1：深度优先搜索

/**
 * 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[] {
    if (!root) return [];
    const res = [];
    const dfs = (res, root, curHeight) => {
        if (curHeight === res.length) {
            res.push(root.val);
        } else {
            res.splice(curHeight, 1, Math.max(res[curHeight], root.val));
        }
        if (root.left) {
            dfs(res, root.left, curHeight + 1);
        }
        if (root.right) {
            dfs(res, root.right, curHeight + 1);
        }
    }
    dfs(res, root, 0);
    return res;
};

2 ) 方案2：广度优先搜索

/**
 * 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[] {
    if (!root) return [];
    const res = [];
    const queue = [root];
    while (queue.length) {
        let len = queue.length;
        let maxVal = -Number.MAX_VALUE;
        while (len > 0) {
            len--;
            const t = queue.shift();
            maxVal = Math.max(maxVal, t.val);
            if (t.left) {
                queue.push(t.left);
            }
            if (t.right) {
                queue.push(t.right);
            }
        }
        res.push(maxVal);
    }
    return res;
};