LeetCode-144. Binary Tree Preorder Traversal

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问题描述

Given the root of a binary tree, return the preorder traversal of its nodes’ values.

Constraints:

  • The number of nodes in the tree is in the range [0, 100].
  • -100 <= Node.val <= 100

解答

方法一：递归
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/*
 * @Description: 144. Binary Tree Preorder 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: 递归写法一
 * @param {TreeNode} root
 * @return {number[]}
 */
const preorderTraversal = function (root) {
  if (!root) {
    return []
  }
  return [root.val, ...preorderTraversal(root.left), ...preorderTraversal(root.right)]
}

/**
 * @description: 递归写法二
 * @param {TreeNode} root
 * @return {number[]}
 */
const preorderTraversal1 = function (root) {
  let res = []
  const visitLoop = (root) => {
    if (root) {
      res.push(root.val)
      visitLoop(root.left)
      visitLoop(root.right)
    }
  }
  visitLoop(root)
  return res
}
方法二：迭代
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/*
 * @Description: 144. Binary Tree Preorder 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: 非递归
 * @param {TreeNode} root
 * @return {number[]}
 */
const preorderTraversal2 = function (root) {
  let stack = []
  let res = []
  let cur = root
  while (cur || stack.length) {
    if (cur) {
      stack.push(cur)
      res.push(cur.val)
      cur = cur.left
    } else {
      cur = stack.pop()
      cur = cur.right
    }
  }
  return res
}