Binary Tree Inorder Traversal

描述

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

For example:

Given binary tree {1,#,2,3},

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

return [1,3,2].

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

分析

用栈或者Morris遍历。

栈

  1. // Binary Tree Inorder 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> inorderTraversal(TreeNode root) {
  5.         ArrayList<Integer> result = new ArrayList<>();
  6.         Stack<TreeNode> s = new Stack<>();
  7.         TreeNode p = root;
  9.         while (!s.empty() || p != null) {
  10.             if (p != null) {
  11.                 s.push(p);
  12.                 p = p.left;
  13.             } else {
  14.                 p = s.pop();
  15.                 result.add(p.val);
  16.                 p = p.right;
  17.             }
  18.         }
  19.         return result;
  20.     }
  21. }

Morris中序遍历

  1. // Binary Tree Inorder Traversal
  2. // Morris中序遍历，时间复杂度O(n)，空间复杂度O(1)
  3. class Solution {
  4.     public List inorderTraversal(TreeNode root) {
  5.         ArrayList result = new ArrayList<>();
  6.         TreeNode cur = root;
  7.         TreeNode prev = null;
  9.         while (cur != null) {
  10.             if (cur.left == null) {
  11.                 result.add(cur.val);
  12.                 prev = cur;
  13.                 cur = cur.right;
  14.             } else {
  15.                 /* 查找前驱 */
  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; 不能有这句，cur还没有被访问 */
  23.                     cur = cur.left;
  24.                 } else {    /* 已经线索化，则访问节点，并删除线索  */
  25.                     result.add(cur.val);
  26.                     node.right = null;
  27.                     prev = cur;
  28.                     cur = cur.right;
  29.                 }
  30.             }
  31.         }
  32.         return result;
  33.     }
  34. }

相关题目

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