Given a binary tree, return all root-to-leaf paths.
Note: A leaf is a node with no children.
Example:
Input:
1
/ \
2 3
\
5
Output: ["1->2->5", "1->3"]
Explanation: All root-to-leaf paths are: 1->2->5, 1->3
/**
* 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 {
List<String> list=new ArrayList<>();
public List<String> binaryTreePaths(TreeNode root) {
if(root == null){
return list;
}
String s = "";
dfs(root,s);
return list;
}
public void dfs(TreeNode root, String s){
if(root.left==null && root.right==null){
list.add(s+root.val);
return;
}
if(root.left!=null){
dfs(root.left, s+root.val+"->");
}
if(root.right!=null){
dfs(root.right, s+root.val+"->");
}
}
}深度优先遍历采用递归
广度优先遍历采用queue