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[LeetCode] #257 二叉樹的所有路徑

[LeetCode] #257 二叉樹的所有路徑

給你一個二叉樹的根節點 root ,按 任意順序 ,返回所有從根節點到葉子節點的路徑。

葉子節點 是指沒有子節點的節點。

輸入:root = [1,2,3,null,5]

輸出:["1->2->5","1->3"]

遞迴,將list作為引數傳遞

class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> list = new ArrayList<>();
        if(root==null) return
list; helper(root, "", list); return list; } public void helper(TreeNode root, String path, List<String> list){ if(root == null) return; if(root.left == null && root.right == null) list.add(path + root.val); else{ helper(root.left, path
+root.val+"->", list); helper(root.right, path+root.val+"->", list); } } }

遞迴,將list作為全域性變數

class Solution {
    List<String> list = new ArrayList<>();

    public List<String> binaryTreePaths(TreeNode root) {
        helper(root, "");
        return list;
    }
    
public void helper(TreeNode root, String path){ if(root == null) return; if(root.left == null && root.right == null) list.add(path + root.val); else{ helper(root.left, path+root.val+"->"); helper(root.right, path+root.val+"->"); } } }

遞迴,將list作為返回值

class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        return helper(root, "");
    }
    public List helper(TreeNode root, String path){
        List<String> list = new ArrayList<>();
        if(root == null) return list;
        if(root.left == null && root.right == null) list.add(path + root.val);
        else{
            list.addAll(helper(root.left, path+root.val+"->"));
            list.addAll(helper(root.right, path+root.val+"->"));
        }
        return list;
    }
}

迭代

class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> paths = new ArrayList<String>();
        if (root == null) return paths;

        Queue<TreeNode> nodeQueue = new LinkedList<TreeNode>();
        Queue<String> pathQueue = new LinkedList<String>();

        nodeQueue.offer(root);
        pathQueue.offer(Integer.toString(root.val));

        while (!nodeQueue.isEmpty()) {
            TreeNode node = nodeQueue.poll(); 
            String path = pathQueue.poll();

            if (node.left == null && node.right == null) {
                paths.add(path);
            } else {
                if (node.left != null) {
                    nodeQueue.offer(node.left);
                    pathQueue.offer(new StringBuffer(path).append("->").append(node.left.val).toString());
                }

                if (node.right != null) {
                    nodeQueue.offer(node.right);
                    pathQueue.offer(new StringBuffer(path).append("->").append(node.right.val).toString());
                }
            }
        }
        return paths;
    }
}

知識點:

總結: