112. 路径总和n 简单 dfs bfs 两种写法
- 路径总和
给你二叉树的根节点 root 和一个表示目标和的整数 targetSum 。判断该树中是否存在 根节点到叶子节点 的路径,这条路径上所有节点值相加等于目标和 targetSum 。如果存在,返回 true ;否则,返回 false 。
叶子节点 是指没有子节点的节点。
示例 1:
输入:root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
输出:true
解释:等于目标和的根节点到叶节点路径如上图所示。
示例 2:
输入:root = [1,2,3], targetSum = 5
输出:false
解释:树中存在两条根节点到叶子节点的路径:
(1 --> 2): 和为 3
(1 --> 3): 和为 4
不存在 sum = 5 的根节点到叶子节点的路径。
示例 3:
输入:root = [], targetSum = 0
输出:false
解释:由于树是空的,所以不存在根节点到叶子节点的路径。
提示:
树中节点的数目在范围 [0, 5000] 内
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
递归
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean hasPathSum(TreeNode root, int targetSum) {
return dfs(root,0,targetSum);
}
boolean dfs(TreeNode root ,int s,int target){
if(root==null)return false;
if(root.left==null&&root.right==null&&s+root.val==target)return true;
boolean flag=dfs(root.left,s+root.val,target)||dfs(root.right,s+root.val,target);
return flag;
}
}
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean hasPathSum(TreeNode root, int targetSum) {
return dfs(root,targetSum);
}
boolean dfs(TreeNode root ,int target){
if(root==null)return false;
if(root.left==null&&root.right==null&&root.val==target)return true;
boolean flag=dfs(root.left,target-root.val)||dfs(root.right,target-root.val);
return flag;
}
}
bfs
bfs开两个队列就可以做到类似于dfs传递参数的效果
import java.util.LinkedList;
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean hasPathSum(TreeNode root, int targetSum) {
Queue<TreeNode> q=new LinkedList<>();
Queue<Integer> sq=new LinkedList<>();
q.offer(root);
sq.offer(0);
while(!q.isEmpty()){
TreeNode t=q.poll();
int s=sq.poll();
if(t==null)continue;
if(t.left==null&&t.right==null&&s+t.val==targetSum)return true;
q.offer(t.left);
sq.offer(s+t.val);
q.offer(t.right);
sq.offer(s+t.val);
}
return false;
}
}