二叉树根节点的所有路径打印

Author: yangjunbao

src/com/apptao/leetcode/tree/BinaryTreePaths.java

@@ -0,0 +1,93 @@
+package com.apptao.leetcode.tree;
+
+import java.util.ArrayList;
+import java.util.List;
+
+/**
+ * 写一个算法，得出二叉树所有根节点到叶子节点的路径
+ * Created by a on 2017-12-28.
+ * <p>
+ * 例如：
+ * 1
+ * /   \
+ * 2     3
+ * \
+ * 5
+ * 结果:
+ * <p>
+ * ["1->2->5", "1->3"]
+ */
+public class BinaryTreePaths {
+
+    public class TreeNode {
+        int val;
+        TreeNode leftChild;
+        TreeNode rightChild;
+
+        TreeNode(int x) {
+            val = x;
+        }
+    }
+
+    /**
+     * 创建一棵二叉树
+     * <pre>
+     *          1
+     *     2         3
+     *  4    5            6
+     *  </pre>
+     */
+    public TreeNode createBinTree() {
+        TreeNode root = new TreeNode(1);
+        TreeNode newNode2 = new TreeNode(2);
+        TreeNode newNode3 = new TreeNode(3);
+        TreeNode newNode4 = new TreeNode(4);
+        TreeNode newNode5 = new TreeNode(5);
+        TreeNode newNode6 = new TreeNode(6);
+        root.leftChild = newNode2;
+        root.rightChild = newNode3;
+        root.leftChild.leftChild = newNode4;
+        root.leftChild.rightChild = newNode5;
+        root.rightChild.rightChild = newNode6;
+        return root;
+    }
+
+    public static void main(String[] args) {
+
+        BinaryTreePaths binaryTreePaths = new BinaryTreePaths();
+
+        List<String> strings = binaryTreePaths.binaryTreePaths(binaryTreePaths.createBinTree());
+
+        System.out.println(strings.toString());
+    }
+
+    /**
+     * 接收数据进行处理，求二叉树的路径
+     * @param root TreeNode
+     * @return list
+     */
+    public List<String> binaryTreePaths(TreeNode root) {
+        List<String> result = new ArrayList<>();
+        new BinaryTreePaths().inorder(root, result, "");
+        return result;
+    }
+
+    /**
+     * 递归路径，添加到list
+     * @param node TreeNode
+     * @param list List<String>
+     * @param path String
+     */
+    private void inorder(TreeNode node, List<String> list, String path) {
+        if (node != null) {
+            if (node.leftChild == null && node.rightChild == null) {
+                list.add(path + node.val);
+            } else {
+                inorder(node.leftChild, list, path + node.val + "->");
+                inorder(node.rightChild, list, path + node.val + "->");
+            }
+        }
+    }
+
+
+}