二叉树的所有路径

九章答案

// version 1: Divide Conquer
public class Solution {
    /**
     * @param root the root of the binary tree
     * @return all root-to-leaf paths
     */
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> paths = new ArrayList<>();
        if (root == null) {
            return paths;
        }
        
        List<String> leftPaths = binaryTreePaths(root.left);
        List<String> rightPaths = binaryTreePaths(root.right);
        for (String path : leftPaths) {
            paths.add(root.val + "->" + path);
        }
        for (String path : rightPaths) {
            paths.add(root.val + "->" + path);
        }
        
        // root is a leaf
        if (paths.size() == 0) {
            paths.add("" + root.val);
        }
        
        return paths;
    }
}

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我写的

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root the root of the binary tree
     * @return all root-to-leaf paths
     */
    public List<String> binaryTreePaths(TreeNode root) {
        // Write your code here
       
             List<String> paths = new ArrayList<String>();
             if(root == null) {
                 return paths;
             }
             //leaf
             if(root.left == null && root.right == null) {
                 paths.add(root.val + "");//不能paths.add(root.val);要转换成string
             }
             List<String> leftpath = binaryTreePaths(root.left);
             List<String> rightpath = binaryTreePaths(root.right);
             
             for(String path : leftpath) {
                 paths.add(root.val + "->" + path);
             }
             for(String path : rightpath) {
                 paths.add(root.val + "->" + path);
             }
             return paths;
    }
             
            
    
}

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转载于:https://www.cnblogs.com/yunyouhua/p/6720233.html