一、写在前面 本篇博客主要是对树的一些遍历方式进行总结,其中包含树的深度优先遍历 ,和树的广度优先遍历 ,以及二叉树的先中后序遍历 ,其中先中后序遍历 分别使用递归版本 的遍历方式和非递归版本 的递归方式。 二、树的深度优先遍历和广度优先遍历 如上图所示为我们需要遍历的数结构。
数据可以整理为:
const tree = {
val: 'a',
children: [{
val: 'b',
children: [{
val: 'd',
children: [],
},
{
val: 'e',
children: [],
}
],
},
{
val: 'c',
children: [{
val: 'f',
children: [],
},
{
val: 'g',
children: [],
}
],
}
],
};
深度优先遍历
const dfs = (root) => {
console.log(root.val)
root.children.forEach(item => {
dfs(item)
})
}
dfs(tree)
广度优先遍历
const bfs = (node) => {
let queue = [node]
while(queue.length > 0) {
let curNode = queue.shift()
console.log(curNode.val)
let child = curNode.children
if(child && child.length > 0) {
child.forEach(item => {
queue.push(item)
})
}
}
}
bfs(tree)
总结 :树的深度优先遍历使用的是递归,树的广度优先遍历使用的是:队列这种数据结构。 三、二叉树的先中后序遍历 database
const dbs = {
val: 4,
left: {
val: 3,
left: {
val: 2,
left: null,
right: null
},
right: {
val: 1,
left: null,
right: null
}
},
right: {
val: 5,
left: {
val: 6,
left: null,
right: null
},
right: {
val: 7,
left: null,
right: null
}
}
}
先序遍历
const preorder = (root) => {
if(!root) return
console.log(root.val)
if(root.left) preorder(root.left)
if(root.right) preorder(root.right)
}
preorder(dbs)
中序遍历
const midorder = (root) => {
if(!root) return
if(root.left) midorder(root.left)
console.log(root.val)
if(root.right) midorder(root.right)
}
midorder(dbs)
后序遍历
const afterorder = (root) => {
if(!root) return
if(root.left) afterorder(root.left)
if(root.right) afterorder(root.right)
console.log(root.val)
}
afterorder(dbs)
上述递归版的遍历方式太简单了,下面是非递归版的写法。 先序遍历——非递归版 :使用数据结构栈
const preorder = (root) => {
if (!root) return
let stack = [root]
while (stack.length > 0) {
let cur = stack.pop()
console.log(cur.val)
if(cur.right) stack.push(cur.right)
if(cur.left) stack.push(cur.left)
}
}
preorder(dbs)
中序遍历——非递归版
const midorder = (root) => {
if(!root) return
let stack = []
let p = root
while(stack.length > 0 || p) {
while(p) {
stack.push(p)
p = p.left
}
let cur = stack.pop()
console.log(cur.val)
p = cur.right
}
}
midorder(dbs)
后序遍历——非递归版
const postorder = (root) => {
if(!root) return
let stack1 = [root]
let stack2 = []
while(stack1.length > 0) {
let cur = stack1.pop()
stack2.push(cur)
if(cur.left) stack1.push(cur.left)
if(cur.right) stack1.push(cur.right)
}
while(stack2.length > 0) {
let cur = stack2.pop()
console.log(cur.val)
}
}
postorder(dbs)
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