Binary Tree Paths 解答

Question

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

For example, given the following binary tree:

   1
 /   
2     3
 
  5

All root-to-leaf paths are:

["1->2->5", "1->3"]

Solution -- Recursive

We can not apply inorder traversal here because it only show path from root to leaf node as required. Therefore, we use recursive way.

We consider two situation:

1. Current node is leaf. Add string to result.

2. Current node has children. Go on.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> result = new ArrayList<String>();
        if (root == null)
            return result;
        dfs(root, result, new StringBuilder());
        return result;
    }
    
    private void dfs(TreeNode root, List<String> result, StringBuilder current) {
        if (root == null)
            return;
        if (root.left == null && root.right == null) {
            current.append(root.val);
            result.add(current.toString());
            return;
        }
        
        current.append(root.val);
        current.append("->");
        
        StringBuilder tmp1 = new StringBuilder(current);
        StringBuilder tmp2 = new StringBuilder(current);
        if (root.left != null)
            dfs(root.left, result, tmp1);
        if (root.right != null)
            dfs(root.right, result, tmp2);
    }
}