Binary Tree Inorder Traversal leetcode java

本文轉載自查看原文 2014-07-30 09:24 2515 leetcode每題整理

題目：

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

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

return [1,3,2].

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

題解：

中序遍歷：遞歸左處理當前遞歸右。

畫圖的話就是，之前離散老師教的，從root開始沿着子樹畫線，遍歷完全樹，每一個被畫線畫到2次的就表示遍歷到了，依次寫出就行了。

遞歸代碼如下：

1      public void helper(TreeNode root, ArrayList<Integer> re){
2          if(root== null)
3              return;
4         helper(root.left,re);
5         re.add(root.val);
6         helper(root.right,re);
7     }
8      public ArrayList<Integer> inorderTraversal(TreeNode root) {
9         ArrayList<Integer> re = new ArrayList<Integer>();
10          if(root== null)
11              return re;
12         helper(root,re);
13          return re;
14     }

非遞歸代碼如下：

1 public ArrayList<Integer> inorderTraversal(TreeNode root) {
2     ArrayList<Integer> res = new ArrayList<Integer>();
3      if(root == null)
4          return res;
5     LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
6      while(root!= null || !stack.isEmpty()){
7          if(root!= null){
8             stack.push(root);
9             root = root.left;
10         } else{
11             root = stack.pop();
12             res.add(root.val);
13             root = root.right;
14         }
15     }
16      return res;
17 }

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