leetcode 257 Binary Tree Paths
z
 2018-02-28 20:19:36  

Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

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   1
 /   \
2     3
 \
  5

All root-to-leaf paths are:

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["1->2->5", "1->3"]
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> result = new ArrayList<String>();
        binaryTreePathsFinder(root, "", result);
        return result;
    }
    public void binaryTreePathsFinder(TreeNode root, String path, List<String> result){
        if (root == null) return ;
        if(root.left == null && root.right == null){
            result.add(path + root.val);
        }
        else if( root.left != null && root.right != null){
            path = path + root.val + "->";
            binaryTreePathsFinder(root.left, path, result);
            binaryTreePathsFinder(root.right, path, result);
        }
        else if(root.left != null && root.right == null){
            path = path + root.val + "->";
            binaryTreePathsFinder(root.left, path, result);
        }
        else{
            path = path + root.val + "->";
            binaryTreePathsFinder(root.right, path, result);
        }
    }
}
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public List<String> binaryTreePaths(TreeNode root) {
    List<String> answer = new ArrayList<String>();
    if (root != null) searchBT(root, "", answer);
    return answer;
}
private void searchBT(TreeNode root, String path, List<String> answer) {
    if (root.left == null && root.right == null) answer.add(path + root.val);
    if (root.left != null) searchBT(root.left, path + root.val + "->", answer);
    if (root.right != null) searchBT(root.right, path + root.val + "->", answer);
}