数据结构与算法之二叉树: LeetCode 107. 二叉树的层序遍历 II (Ts版)

二叉树的层序遍历 II

• https://leetcode.cn/problems/binary-tree-level-order-traversal-ii/description/

描述

• 给你二叉树的根节点 root ，返回其节点值 自底向上的层序遍历 。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

示例 1

输入：root = [3,9,20,null,null,15,7]
输出：[[15,7],[9,20],[3]]

示例 2

输入：root = [1]
输出：[[1]]

示例 3

输入：root = []
输出：[]

提示

• 树中节点数目在范围 [0, 2000] 内
• -1000 <= Node.val <= 1000

Typescript 版算法实现

1 ) 方案1: 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)
 *     }
 * }
 */

/**
 * 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 levelOrderBottom(root: TreeNode | null): number[][] {
    let levelOrder: number[][] = [];
    if (root === null) return levelOrder;
    let queue: TreeNode[] = [root];
    while (queue.length > 0) {
        let level: number[] = [];
        let size = queue.length;
        for (let i = 0; i < size; i++) {
            let node = queue.shift()!;
            level.push(node.val);
            if (node.left !== null) queue.push(node.left);
            if (node.right !== null) queue.push(node.right);
        }
        levelOrder.unshift(level); // Insert at the beginning to reverse order.
    }
    return levelOrder;
}

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 dfs(node: TreeNode | null, level: number, levels: number[][]): void {
    if (node === null) return;
    if (level === levels.length) levels.push([]);
    levels[level].push(node.val);
    dfs(node.left, level + 1, levels);
    dfs(node.right, level + 1, levels);
}

function levelOrderBottom(root: TreeNode | null): number[][] {
    const levels: number[][] = [];
    dfs(root, 0, levels);
    return levels.reverse();
}