[leetcode]257.Binary Tree Paths

题目

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

解法一

思路

用数的DFS（深度优先遍历包括先序，中序，后序遍历）。用两个栈，其中一个栈来保存结点，另外一个栈来保存从根节点到当前结点的路径。两个栈一直是同步的，也就是说每次出栈的结点与字符串是对应关系。

代码

/**
 * 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> list = new ArrayList<>();
        Stack<TreeNode> nStack = new Stack<>();
        Stack<String> sStack = new Stack<>();
        if(root == null) return list;
        nStack.push(root);
        sStack.push("");
        
        while(!nStack.isEmpty()){
            TreeNode curNode = nStack.pop();
            String curString = sStack.pop();
            
            if(curNode.left == null && curNode.right == null)
                list.add(curString +  curNode.val);
            if(curNode.right != null) {
                nStack.push(curNode.right);
                sStack.push(curString + curNode.val + "->");
            }
            if(curNode.left != null) {
                nStack.push(curNode.left);
                sStack.push(curString + curNode.val + "->");
            }
        }
        
        return list;
    }
}

解法二

思路

用树的BFS（实际就是层次遍历）。和解法一一样，把栈改成队列即可。

代码

/**
 * 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> list = new ArrayList<>();
        Queue<TreeNode> nQueue = new LinkedList<>();
        Queue<String> sQueue = new LinkedList<>();
        if(root == null) return list;
        nQueue.add(root);
        sQueue.add("");
        
        while(!nQueue.isEmpty()){
            TreeNode curNode = nQueue.remove();
            String curString = sQueue.remove();
            
            if(curNode.left == null && curNode.right == null)
                list.add(curString +  curNode.val);
            if(curNode.right != null) {
                nQueue.add(curNode.right);
                sQueue.add(curString + curNode.val + "->");
            }
            if(curNode.left != null) {
                nQueue.add(curNode.left);
                sQueue.add(curString + curNode.val + "->");
            }
        }
        
        return list;
    }
}

解法三

思路

用递归,其实也是DFS的思想

代码

class Solution {
    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);
    }
}