使用 Javascript 的二叉树级顺序遍历

Question

这是一道leetcode题。

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example: Given binary tree [3, 9, 20, null, null, 15, 7],
    3
   / \
  9  20
    /  \
   15   7
return its level order traversal as:
[
  [3],
  [9,20],
  [15,7]
]

但我正在 JavaScript 中尝试一种新方法，而不是完全按照他们的解决方案进行。到目前为止，我能够打印数组，但是

How can different levels be printed in new rows

下面是我目前的代码：

var levelOrder = function(root) {
let output = [];
let queue = [];
let currentNode = root;
queue.push(currentNode);
let currentLevel = 1;
while(queue.length){
    
    currentNode = queue.shift();
    currentLevel--; //this will ensure we are adding new lines only on next level
    output.push(currentNode);
    
    if(currentNode.left){
        queue.push(currentNode.left);
    }
    if(currentNode.right){
        queue.push(currentNode.right);
    }
    
    if(currentLevel = 0){
        output = output + '/n'; //Insert a new line
        currentLevel = queue.length; //2
    }
}
return output;
};

Input: [3,9,20,null,null,15,7],

Expected Output:
[
[3],
[9,20],
[15,7]
]

LeetCode 问题 Link: BinaryTreeTraversalUsingBFS

Answer 1

我想你快到了。虽然不确定 output = output + '/n'; 是什么。

这会通过：

var levelOrder = function(root) {
    const levels = []

    if(!root) {
        return levels
    }

    const queue = [root]
    while (queue.length){
       const queueLength = queue.length
       const level = []

       for(let i = 0; i < queueLength; i++){

           const node = queue.shift()

           if(node.left){
               queue.push(node.left)
           }
           if(node.right){
               queue.push(node.right)
           }

           level.push(node.val)
       }
       levels.push(level)
   }
    return levels
}

参考资料

有关其他详细信息，您可以在其中查看 Discussion Board. There are plenty of accepted solutions with a variety of languages and explanations, efficient algorithms, as well as asymptotic time/space complexity analysis^{1, 2}。

Answer 2

根据你的代码库，我修改它以供工作。

添加索引以增加输出索引。
使用严格的相等运算符而不是赋值变量。
删除output = output + '/n'，因为输出是一个数组。

var levelOrder = function (root) {
  let output = [];
  let queue = [];
  let currentNode = root;
  queue.push(currentNode);
  let currentLevel = 1;
  let index = 0; // Add an index for increasing the output index

  while (queue.length) {

    currentNode = queue.shift();
    currentLevel--;

    if (!output[index]) { // Set default is an array for each output element in first time
      output[index] = [];
    }

    output[index].push(currentNode.val);

    if (currentNode.left) {
      queue.push(currentNode.left);
    }

    if (currentNode.right) {
      queue.push(currentNode.right);
    }

    if (currentLevel === 0) { // Use strict equality operator to compare 0
      index++; // increase index
      currentLevel = queue.length;
    }
  }

  return output;
};

Answer 3

这是它的递归解法

    var levelOrder = function(root) {
    let result =[];
    if(!root)return result;
    if(!root.left && !root.right){
        result.push([root.val]);
        return result;
    }
    const pushIntoResult =(node, level) =>{
        if(!node) return;
        if(!result[level]){
            result.push([]);
        }
        result[level].push(node.val);
        pushIntoResult(node.left, level+1);
        pushIntoResult(node.right, level+1);
    }
    pushIntoResult(root, 0);
    return result;
};

Answer 4

您可以在@Ahmad Atrach 的代码中尝试这个位修改

const dfs = (node,result,level) => {

if(!node) return;
if(!result[level]){
    result.push([]);
}
result[level].push(node.val);
dfs(node.left,result, level+1);
dfs(node.right,result, level+1);
}
var levelOrder = function(root) {

  if(!root) return [];
  let result = [];
  dfs(root,result,0);

  return result;
};

使用 Javascript 的二叉树级顺序遍历

Binary Tree Level Order Traversal using Javascript

javascript

binary-tree

breadth-first-search

multidimensional-array

参考资料