LeetCode解题报告:Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

注意：下面是迭代的解法。理解有点困难，和大家讨论一下。

 1 import java.util.ArrayList;
 2 import java.util.List;
 3 import java.util.Stack;
 4
 5 /**
 6  * Definition for binary tree public class TreeNode { int val; TreeNode left;
 7  * TreeNode right; TreeNode(int x) { val = x; } }
 8  */
 9 public class TreeNodeNoneReverse {
10     List<Integer> postOrder = new ArrayList<Integer>();
11
12     public List<Integer> postorderTraversal(TreeNode root) {
13         // 非递归实现后序遍历
14         TreeNode tempTree = root;
15         Stack<TreeNode> stack = new Stack<TreeNode>();
16         while (root != null) {
17             // 左子树入栈
18             for (; root.left != null; root = root.left) {
19                 System.out.println("左子树入栈：" + root.val);
20                 stack.push(root);
21             }
22             // 当前节点无右子或右子已经输出
23             while (root != null
24                     && (root.right == null || root.right == tempTree)) {
25                 System.out.println("add list：" + root.val);
26                 postOrder.add(root.val);
27                 tempTree = root;// 记录上一个已输出节点
28                 System.out.println("上一个已输出的节点："+tempTree.val);
29                 if (stack.empty()) {
30                     System.out.println("stack is empty.");
31                     return postOrder;
32                 }
33                 root = stack.pop();
34                 System.out.println("右子树出栈：" + root.val);
35             }
36             // 处理右子
37             System.out.println("处理右子入栈：" + root.val);
38             stack.push(root);
39             root = root.right;
40         }
41         System.out.println("可能运行到这儿么？");
42         return postOrder;
43     }
44
45     public static void main(String[] args) {
46         TreeNode t1 = new TreeNode(1);
47         TreeNode t2 = new TreeNode(2);
48         TreeNode t3 = new TreeNode(3);
49         TreeNode t4 = new TreeNode(4);
50         TreeNode t5 = new TreeNode(5);
51         TreeNode t6 = new TreeNode(6);
52         t3.setLeft(t2);
53         t3.setRight(t1);
54         t2.setLeft(t4);
55         t2.setRight(t5);
56         t1.setRight(t6);
57         System.out.println(new TreeNodeNoneReverse().postorderTraversal(t3));
58     }
59 }
60 　　

LeetCode解题报告:Binary Tree Postorder Traversal