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路径总和 II-113 #27
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路径总和 II-113 #27
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|
data.json {
"code": 1,
"children": [
{
"code": 2,
"children": [
{
"code": 5,
"children": [
{
"code": 7
},
{
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}
]
}
]
},
{
"code": 3,
"children": [
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{
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}
]
},
{
"code": 90,
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},
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},
{
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},
{
"code": 16,
"children": [
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"code": 74,
"children": [
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"code": 70,
"children": [
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{
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"children": [
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"code": 888
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{
"code": 999,
"children": [
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]
},
{
"code": 555
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]
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}
]
},
{
"code": 4,
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}index.js const data = require('./data.json')
function isChildren(node){
return node.children?true:false
}
var pathCode = function(node, code,key) {
const pathes = [];
_pathCode(node, code, pathes, [],key);
return pathes;
};
function _pathCode(node, code, pathes, path,key) {
path = [...path, node[key]];
if (node[key]===code){
pathes.push(...path)
return
}
if (isChildren(node)){
for (let i=0;i<node.children.length;i++){
_pathCode(node.children[i],code,pathes,path,key)
}
}
}
const paths = pathCode(data,555,'code') |
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给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。
说明: 叶子节点是指没有子节点的节点。
示例:
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/path-sum-ii
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
思路
典型的可以用 DFS 来解决的问题，定义一个 search 方法并且参数里带一个用来收集路径的 paths 数组，每当到达叶子节点（没有 left 也没有 right），就计算一把路径的总和，如果等于目标值就 push 到结果数组里。（注意这里要浅拷贝一下，防止下面的计算污染这个数组）
任何一个节点处理完成时，都要把当前节点 pop 出 paths 数组。