Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:Given the below binary tree and
sum = 22
,
5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
題意:
給定一個二叉樹,每個節點都有一個數字,還有一個目標數。找到一條路徑加總起來有達到目標數則返回路徑。
題解:
對這棵二叉樹進行左右子樹的歷遍,若到子節點有達到目標數則添加這個路徑。
*注意偏斜樹*
package LeetCode.Medium;
import java.util.ArrayList;
import java.util.List;
import LeetCode.Dependencies.TreeNode;
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class PathSumII {
List<List<Integer>> result = new ArrayList<>();
public List<List<Integer>> pathSum(TreeNode root, int sum) {
if(root == null)
return result;
List<Integer> temp = new ArrayList<>();
temp.add(root.val);
dfs(root, sum - root.val, temp);
return result;
}
void dfs(TreeNode root, int sum, List<Integer> temp) {
if(root.left == null && root.right == null) {
if(sum == 0) {
List<Integer> subResult = new ArrayList<Integer>(temp);
result.add(subResult);
}
return;
}
if(root.left != null) {
temp.add(root.left.val);
//注意sum不能影響到下一個狀態
int nextSum = sum - root.left.val;
dfs(root.left, nextSum, temp);
//回復上一個狀態
temp.remove(temp.size() - 1);
}
if(root.right != null) {
temp.add(root.right.val);
int nextSum = sum - root.right.val;
dfs(root.right, nextSum, temp);
//回復上一個狀態
temp.remove(temp.size() - 1);
}
}
}