Problem: 257. 二叉树的所有路径

题目描述

思路

遍历思想(利用二叉树的先序遍历)

利用先序遍历的思想，我门用一个List变量path记录当前先序遍历的节点，当遍历到根节点时，将其添加到另一个List变量res中，当递归往回归的时候删除当前path中的最后一个值

复杂度

时间复杂度:

$O(n)$;其中$n$为二叉树的节点个数

空间复杂度:

$O(h)$；其中$h$为二叉树的高度

Code

/*
/**
 * 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 {
    public List<String> binaryTreePaths(TreeNode root) {
        traverse(root);
        return res;
    }

    // Record the traverse recursive path
    LinkedList<String> path = new LinkedList<>();
    // Records all paths from the root to the leaf node
    LinkedList<String> res = new LinkedList<>();

    private void traverse(TreeNode root) {
        if (root == null) {
            return;
        }
        // leaf root
        if (root.left == null && root.right == null) {
            path.addLast(root.val + "");
            // Add this path to res
            res.addLast(String.join("->", path));
            path.removeLast();
            return;
        }
        // Preorder traversal position
        path.addLast(root.val + "");
        // Recursively traverse the left and right subtrees
        traverse(root.left);
        traverse(root.right);
        // Post order traversal position
        path.removeLast();
    }
}