代码随想录刷题记录-Day14

发表于 本文字数： 495 阅读时长 ≈ 2 分钟

To Do List

  • DAY 13 回顾
  • 二叉树的理论知识的理解
  • 二叉树的递归遍历三部曲
  • DAY 14 总结

DAY 11 回顾

（）

二叉树的理论知识的理解

（带补充）

二叉树的递归遍历三部曲（前序、中序、后序）

前序遍历

1
Difficulty: Easy

给你二叉树的根节点 root ，返回它节点值的 前序 遍历

Example 1:

img

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Input: root = [1,null,2,3]
Output: [1,2,3]

Example 2:

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Input: root = []
Output: []

Example 3:

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Input: root = [1]
Output: [1]

Constraints:

  • The number of nodes in the tree is in the range [0, 100].

  • -100 <= Node.val <= 100

    Translated by zh-CN 树中节点数目在范围 [0, 100] 内

解法：

(JavaScript)

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/**
 * 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[]}
 */
var preorderTraversal = function(root) {
    let res = []
    let recur = function(root){
        if (root == null) return
        res.push(root.val)
        recur(root.left)
        recur(root.right)
    }
    recur(root)
    return res
};

PS:

后序遍历

1
Difficulty: Easy

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

Example 1:

img

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Input: root = [1,null,2,3]
Output: [3,2,1]

Example 2:

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Input: root = []
Output: []

Example 3:

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Input: root = [1]
Output: [1]

Constraints:

  • The number of the nodes in the tree is in the range [0, 100].

  • -100 <= Node.val <= 100

    Translated by zh-CN 树中节点数目在范围 [0, 100] 内

解法：

(JavaScript)

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/**
 * 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[]}
 */
var postorderTraversal = function(root) {
    let res = []
    let recur = function(root){
        if (root == null) return
        recur(root.left)
        recur(root.right)
        res.push(root.val)
    }
    recur(root)
    return res
};

PS:

中序遍历

1
Difficulty: Easy

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

Example 1:

img

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Input: root = [1,null,2,3]
Output: [3,2,1]

Example 2:

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Input: root = []
Output: []

Example 3:

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Input: root = [1]
Output: [1]

Constraints:

  • The number of the nodes in the tree is in the range [0, 100].
  • -100 <= Node.val <= 100
Translated by zh-CN 树中节点数目在范围 [0, 100] 内
---

解法：

(JavaScript)

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/**
 * 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[]}
 */
var inorderTraversal = function(root) {
    let res = []
    let recur = function(root){
        if (root == null) return
        recur(root.left)
        res.push(root.val)
        recur(root.right)
    }
    recur(root)
    return res
};