LeetCode-102. Binary Tree Level Order Traversal

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问题描述

Given the root of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level).

Example :

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Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]

解答

方法一：迭代
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/*
 * @Description: 102. Binary Tree Level Order Traversal
 * @Author: libk
 * @Github: https://github.com/libk
 */
/**
 * 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[][]}
 */
const levelOrder = function(root) {
  if (!root) {
    return []
  }
  let res = []
  let queue = [root]

  while (queue.length) {
    let curLength = queue.length
    let curLevel = []
    for (let i = 0; i < curLength; i++) {
      let cur = queue.shift()
      curLevel.push(cur.val)
      if (cur.left) {
        queue.push(cur.left)
      }
      if (cur.right) {
        queue.push(cur.right)
      }
    }
    res.push(curLevel)
  }
  return res
}

方法二：递归
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/*
 * @Description: 102. Binary Tree Level Order Traversal
 * @Author: libk
 * @Github: https://github.com/libk
 */
/**
 * 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)
 * }
 */
/**
 * @description: 递归
 * @param {TreeNode} root
 * @return {number[][]}
 */
const levelOrder = function(root) {
  let res = []
  let level = 0
  const eventLoop = (cur, level) => {
    if (cur) {
      // 如果当前层未被创建过，则创建当前层
      if (level >= res.length) {
        res[level] = []
      }
      res[level].push(cur.val)
      eventLoop(cur.left, level + 1)
      eventLoop(cur.right, level + 1)
    }
  }
  eventLoop(root, level)
  return res
}