Data Structure: Tree & tree generator All In One

"use strict";

/**
 *
 * @author xgqfrms
 * @license MIT
 * @copyright xgqfrms
 * @created 2022-10-10
 * @modified
 *
 * @description树 & 遍历
 * @difficulty Easy
 * @ime_complexity O(n)
 * @space_complexity O(n)
 * @augments
 * @example
 * @link https://leetcode.cn/leetbook/read/data-structure-binary-tree/xe17x7/
 * @solutions
 *
 * @best_solutions
 *
 */

// export {};

const log = console.log;

// 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);
  }
}


// bug ❌
const treeGenerator = (list) => {
  if(!list.length) return [];
  // let tree = new TreeNode(list[0]);
  let tree = new TreeNode();
  let temp = tree;
  console.log(`tree =`, tree);
  for (const [i, item] of list.entries()) {
    // console.log(`item =`, item);
    // if(i === 0) {
    //   continue;
    // }
    temp.val = item;
    console.log(`temp 1 =`, temp);
    if(item !== null) {
      temp.right = new TreeNode(null);
      temp.left = new TreeNode(list[i + 1]);
      if(i < list.length - 1) {
          temp = temp.left;
      }
    } else {
      temp.left = new TreeNode(null);
      temp.right = new TreeNode(list[i + 1]);
      if(i < list.length - 1) {
        temp = temp.right;
      }
    }
    console.log(`temp 2 =`, temp);
  }
  console.log(`tree ?=`, tree);
  return tree;
}

treeGenerator([1,null,2,3]);
// 

https://leetcode.cn/leetbook/read/data-structure-binary-tree/xecaj6/

LeetCode

function preOrderTraversal(root: TreeNode | null): number[] {
  if(!root) return [];
  // 144. 二叉树的前序遍历: root, left, right
  // DFS 深度优先搜索
  function dfs(head: TreeNode | null, result: number[]) {
    if(!head) {
      return;
    }
    result.push(head.val);
    dfs(head.left, result)
    dfs(head.right, result);
  }
  let result: number[] = [];
  dfs(root, result);
  return result;
};

https://leetcode.com/problems/binary-tree-preorder-traversal/

https://leetcode.cn/problems/binary-tree-preorder-traversal/

refs

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标签：right,TreeNode,generator,temp,tree,Tree,null,Data,left
From： https://www.cnblogs.com/xgqfrms/p/16803069.html