Binary Tree Postorder Traversal

描述

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

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

  1.  1
  2.   \
  3.    2
  4.   /
  5.  3

return [3,2,1].

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

分析

用栈或者Morris遍历。

栈

  1. // Binary Tree Postorder Traversal
  2. // 使用栈，时间复杂度O(n)，空间复杂度O(n)
  3. class Solution {
  4.     public List<Integer> postorderTraversal(TreeNode root) {
  5.         ArrayList<Integer> result = new ArrayList<>();
  6.         Stack<TreeNode> s = new Stack<>();
  7.         /* p，正在访问的结点，q，刚刚访问过的结点*/
  8.         TreeNode p = root;
  9.         TreeNode q = null;
  11.         do {
  12.             while (p != null) { /* 往左下走*/
  13.                 s.push(p);
  14.                 p = p.left;
  15.             }
  16.             q = null;
  17.             while (!s.empty()) {
  18.                 p = s.pop();
  19.                 /* 右孩子不存在或已被访问，访问之*/
  20.                 if (p.right == q) {
  21.                     result.add(p.val);
  22.                     q = p; /* 保存刚访问过的结点*/
  23.                 } else {
  24.                     /* 当前结点不能访问，需第二次进栈*/
  25.                     s.push(p);
  26.                     /* 先处理右子树*/
  27.                     p = p.right;
  28.                     break;
  29.                 }
  30.             }
  31.         } while (!s.empty());
  33.         return result;
  34.     }
  35. }

Morris后序遍历

  1. // Binary Tree Postorder Traversal
  2. // Morris后序遍历，时间复杂度O(n)，空间复杂度O(1)
  3. class Solution {
  4.     public List postorderTraversal(TreeNode root) {
  5.         ArrayList result = new ArrayList<>();
  6.         TreeNode dummy = new TreeNode(-1);
  7.         dummy.left = root;
  8.         TreeNode cur = dummy;
  9.         TreeNode prev = null;
  11.         while (cur != null) {
  12.             if (cur.left == null) {
  13.                 prev = cur; /* 必须要有 */
  14.                 cur = cur.right;
  15.             } else {
  16.                 TreeNode node = cur.left;
  17.                 while (node.right != null && node.right != cur)
  18.                     node = node.right;
  20.                 if (node.right == null) { /* 还没线索化，则建立线索 */
  21.                     node.right = cur;
  22.                     prev = cur; /* 必须要有 */
  23.                     cur = cur.left;
  24.                 } else { /* 已经线索化，则访问节点，并删除线索  */
  25.                     visit_reverse(cur.left, prev, result);
  26.                     prev.right = null;
  27.                     prev = cur; /* 必须要有 */
  28.                     cur = cur.right;
  29.                 }
  30.             }
  31.         }
  32.         return result;
  33.     }
  34.     // 逆转路径
  35.     private static void reverse(TreeNode from, TreeNode to) {
  36.         TreeNode x = from;
  37.         TreeNode y = from.right;
  38.         TreeNode z = null;
  39.         if (from == to) return;
  41.         while (x != to) {
  42.             z = y.right;
  43.             y.right = x;
  44.             x = y;
  45.             y = z;
  46.         }
  47.     }
  49.     // 访问逆转后的路径上的所有结点
  50.     private static void visit_reverse(TreeNode from, TreeNode to,
  51.                                       List result) {
  52.         TreeNode p = to;
  53.         reverse(from, to);
  55.         while (true) {
  56.             result.add(p.val);
  57.             if (p == from)
  58.                 break;
  59.             p = p.right;
  60.         }
  62.         reverse(to, from);
  63.     }
  64. }

相关题目

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/binary-tree/traversal/binary-tree-postorder-traversal.html