113. Path Sum II
题目 113. Path Sum II
思路分析
dfs
代码实现
/**
* 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 List<List<Integer>> pathSum(TreeNode root, int targetSum) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> path = new ArrayList<>();
dfs(root,targetSum,path,result);
return result;
}
private void dfs(TreeNode node,int targetSum,List<Integer> path,List<List<Integer>> result){
if(node == null) return;
targetSum-=node.val;
path.add(node.val);
if(node.left == null && node.right == null){
if(targetSum==0){
result.add(new ArrayList<>(path));
}
}else{
dfs(node.left,targetSum,path,result);
dfs(node.right,targetSum,path,result);
}
targetSum+=node.val;
path.remove(path.size()-1);
}
}
/**
* 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 List<List<Integer>> pathSum(TreeNode root, int targetSum) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> path = new ArrayList<>();
dfs(root,targetSum,path,result);
return result;
}
private void dfs(TreeNode node,int targetSum,List<Integer> path,List<List<Integer>> result){
if(node == null) return;
path.add(node.val);
if(node.left == null && node.right == null){
if(targetSum==node.val){
result.add(new ArrayList<>(path));
}
}else{
dfs(node.left,targetSum-node.val,path,result);
dfs(node.right,targetSum-node.val,path,result);
}
path.remove(path.size()-1);
}
}
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