輸入一棵二叉樹和一個整數,列印出二叉樹中節點值的和為輸入整數的所有路徑。從樹的根節點開始往下一直到葉節點所經過的節點形成一條路徑。
示例:
給定如下二叉樹,以及目标和 sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
傳回:
[
[5,4,11,2],
[5,8,4,5]
]
方法一:遞歸
時間複雜度:O(n)
空間複雜度:O(n)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
List<Integer> res = new ArrayList();
public List<Integer> inorderTraversal(TreeNode root) {
if(root==null) return res;
helper(root);
return res;
}
//構造輔助函數
private void helper(TreeNode node){
if(node==null) return ;
helper(node.left);
res.add(node.val);
helper(node.right);
}
}
方法二:疊代
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
//疊代
List<Integer> res = new ArrayList();
//構造輔助棧
Stack<TreeNode> stack = new Stack();
TreeNode cur = root;
while(cur!=null||!stack.isEmpty()){
//周遊左節點
while(cur!=null){
stack.push(cur);
cur = cur.left;
}
cur = stack.pop();
res.add(cur.val);
cur = cur.right;
}
return res;
}
}