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  • 遍历树,找出所有叶子路径

    一、示例:

    树的结构:

    示例中自己构建了图片中的这棵树:

    树节点模型:

    public class TreeNode {
        String value;
        List<TreeNode> children;
    
        public TreeNode() {
            children = new ArrayList<>();
        }
    
        public TreeNode(String value) {
            this.value = value;
            children = new ArrayList<>();
        }
    
        @Override
        public String toString() {
            // TODO Auto-generated method stub
            return value;
        }
    }

    构建树:

    // 创建一棵树
    TreeNode root = new TreeNode("A");
    // 第二层
    root.children.add(new TreeNode("B"));
    root.children.add(new TreeNode("C"));
    // 第三层
    root.children.get(0).children.add(new TreeNode("D"));
    root.children.get(0).children.add(new TreeNode("E"));
    root.children.get(1).children.add(new TreeNode("F"));
    root.children.get(1).children.add(new TreeNode("H"));
    root.children.get(1).children.add(new TreeNode("G"));
    // 第四层
    root.children.get(0).children.get(1).children.add(new TreeNode("I"));

    二、遍历方式

    提供三种方式进行遍历:

    ① 递归形式的深度优先遍历:

    /**
         * 深度优先遍历(递归方式) --- 树(Tree)
         */
        public void recurTree(TreeNode root) {
            List<List<String>> result = new ArrayList<>();
            List<String> path = new ArrayList<>();
            path.add(root.value);
            findPath(result, root, path);
            System.out.println(result);
        }
    
        private void findPath(List<List<String>> result, TreeNode node, List<String> path) {
            if (node.children == null || node.children.size() <= 0) {
                result.add(path);
                return;
            }
    
            for (int i = 0; i < node.children.size(); i++) {
                TreeNode child = node.children.get(i);
                List<String> cPath = new ArrayList<>();
                cPath.addAll(path);
                cPath.add(child.value);
                findPath(result, child, cPath, target);
            }
        }

    ② 非递归的深度优先遍历

    /**
         * 深度优先遍历(非递归方式) ----- 查找树的所有叶子路径
         * 
         * @param root
         *            根节点
         * @return 叶子路径的集合
         */
        public List<List<TreeNode>> dfsTree(TreeNode root) {
            Stack<TreeNode> nodeStack = new Stack<>();
            Stack<List<TreeNode>> pathStack = new Stack<>();
            List<List<TreeNode>> result = new ArrayList<>();
            nodeStack.push(root);
            ArrayList<TreeNode> arrayList = new ArrayList<>();
            arrayList.add(root);
            pathStack.push(arrayList);
    
            while (!nodeStack.isEmpty()) {
                TreeNode curNode = nodeStack.pop();
                List<TreeNode> curPath = pathStack.pop();
    
                if (curNode.children == null || curNode.children.size() <= 0) {
                    result.add(curPath);
                } else {
                    int childSize = curNode.children.size();
                    for (int i = childSize - 1; i >= 0; i--) {
                        TreeNode node = curNode.children.get(i);
                        nodeStack.push(node);
                        List<TreeNode> list = new ArrayList<>(curPath);
                        list.add(node);
                        pathStack.push(list);
                    }
                }
            }
            return result;
        }

    3. 广度优先遍历,遍历所有叶子路径

    /**
         * 广度优先遍历 ---- 查找树的所有叶子路径
         * 
         * @param root
         *            根节点
         * @return 叶子路径的集合
         */
        public List<List<TreeNode>> bfsTree(TreeNode root) {
            Queue<TreeNode> nodeQueue = new LinkedList<>();
            Queue<List<TreeNode>> qstr = new LinkedList<>();
            List<List<TreeNode>> result = new ArrayList<>();
            nodeQueue.add(root);
            ArrayList<TreeNode> arrayList = new ArrayList<>();
            qstr.add(arrayList);
    
            while (!nodeQueue.isEmpty()) {
                TreeNode curNode = nodeQueue.remove();
                List<TreeNode> curList = qstr.remove();
    
                if (curNode.children == null || curNode.children.size() <= 0) {
                    curList.add(curNode);
                    result.add(curList);
                } else {
                    for (int i = 0; i < curNode.children.size(); i++) {
                        TreeNode treeNode = curNode.children.get(i);
                        nodeQueue.add(treeNode);
                        List<TreeNode> list = new ArrayList<>(curList);
                        list.add(curNode);
                        qstr.add(list);
                    }
                }
            }
    
            return result;
        }

    三种方式的输出:

    深度优先遍历(递归):[[A, B, D], [A, B, E, I], [A, C, F], [A, C, H], [A, C, G]]
    广度优先遍历:[[A, B, D], [A, C, F], [A, C, H], [A, C, G], [A, B, E, I]]
    深度优先遍历(非递归):[[A, B, D], [A, B, E, I], [A, C, F], [A, C, H], [A, C, G]]

    三、总结

      示例是查找树的所有叶子节点,举一反三,如果我们是查找树中满足某个条件的路径,也是非常容易了。比如下面中查找 “ E ” 的分支:

    public void recurTree(TreeNode root) {
            List<List<String>> result = new ArrayList<>();
            List<String> path = new ArrayList<>();
            path.add(root.value);
            findPath(result, root, path, "E");
            System.out.println(result);
        }
    
        private void findPath(List<List<String>> result, TreeNode node, List<String> path, String target) {
            if (target.equals(node.value)) {
                result.add(path);
                return;
            }
    
            if (node.children == null || node.children.size() <= 0) {
                return;
            }
    
            for (int i = 0; i < node.children.size(); i++) {
                TreeNode child = node.children.get(i);
                List<String> cPath = new ArrayList<>();
                cPath.addAll(path);
                cPath.add(child.value);
                if (result.size() > 0)
                    break;
                findPath(result, child, cPath, target);
            }
        }

    输出:

    [[A, B, E]]
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  • 原文地址:https://www.cnblogs.com/aimqqroad-13/p/10238093.html
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