二叉树的先序遍历

Author: DmrfCoder

LeetCode/Doc/先序遍历二叉树.md

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+# binary-tree-preorder-traversal（先序遍历二叉树）
+
+<center>知识点：树</center>
+
+
+## 题目描述
+
+Given a binary tree, return the postorder traversal of its nodes' values.
+
+For example:
+Given binary tree{1,#,2,3},
+
+![image-20190413113855000](https://ws2.sinaimg.cn/large/006tNc79gy1g20usuv6s0j308y0860so.jpg)
+
+return[1,2,3].
+
+Note: Recursive solution is trivial, could you do it iteratively?
+
+
+给定一个二叉树，返回其先序遍历结果。
+
+提示：递归解决方案是不值一提的，你选择迭代。
+
+
+## 解题思路
+
+首先明确什么是先序遍历：对于二叉树访问顺序为根-左-右，即为先序遍历。
+
+递归方式很简单，这里说一下迭代的思路，考虑如下二叉树：
+
+![image-20190413125303689](https://ws1.sinaimg.cn/large/006tNc79gy1g20wxyvib7j30n80ggmyz.jpg)
+使用栈，入栈顺序为根、左，右。
+
+
+## 代码
+
+[这里](../src/seven/Solution.java)

LeetCode/out/production/LeetCode/seven/Solution.class

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+package seven;
+
+/**
+ * @author dmrfcoder
+ * @date 2019-04-13
+ */
+
+import java.util.ArrayList;
+import java.util.Stack;
+
+/**
+ * Definition for binary tree
+ */
+
+class TreeNode {
+    int val;
+    TreeNode left;
+    TreeNode right;
+
+    TreeNode(int x) {
+        val = x;
+    }
+}
+
+public class Solution {
+    public ArrayList<Integer> preorderTraversal(TreeNode root) {
+        Stack<TreeNode> stack = new Stack<>();
+        ArrayList<Integer> arrayList = new ArrayList<>();
+        if (root == null) {
+            return arrayList;
+        }
+
+
+        TreeNode curNode = root;
+        arrayList.add(root.val);
+        stack.push(root);
+
+        while (!stack.empty()) {
+            if (curNode == null) {
+                curNode = stack.pop();
+                if (curNode.right != null) {
+                    arrayList.add(curNode.right.val);
+                    stack.push(curNode.right);
+                    curNode = curNode.right;
+                } else {
+                    curNode = null;
+                }
+                continue;
+            } else if (curNode.left != null) {
+                arrayList.add(curNode.left.val);
+                stack.push(curNode.left);
+
+
+            }
+            curNode = curNode.left;
+        }
+
+        return arrayList;
+
+    }
+
+    public static void main(String args[]) {
+        TreeNode root = new TreeNode(1);
+        TreeNode node2 = new TreeNode(2);
+        TreeNode node3 = new TreeNode(3);
+        TreeNode node4 = new TreeNode(4);
+        TreeNode node5 = new TreeNode(5);
+        TreeNode node6 = new TreeNode(6);
+        TreeNode node7 = new TreeNode(7);
+
+        root.left = node2;
+//        root.right = node3;
+//        node2.left = node4;
+//        node2.right = node5;
+//        node3.left = node6;
+//        node3.right = node7;
+
+        Solution solution = new Solution();
+        ArrayList<Integer> arrayList = solution.preorderTraversal(root);
+        System.out.println(arrayList.toArray().toString());
+
+    }
+}