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Given a binary tree, return the inorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3},
1
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
[Thoughts]
先根遍历的迭代算法比较简单，但是中根遍历的迭代算法稍微难一点。同样借助于栈，但不同于先根遍历的中规中矩，中根遍历需要先把当前结点的所有左结点放入栈中，然后从栈中弹出前一个左结点(即当前的根结点)，把值放入队列，然后指向右结点，重复直至栈空。
[Code]
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<Integer> inorderTraversal(TreeNode root) { ArrayList<Integer> result = new ArrayList<Integer>(); if(root == null){ return result; } Stack<TreeNode> stack = new Stack<TreeNode>(); TreeNode cur = root; while(!stack.isEmpty() || cur != null){ while(cur != null){// put all the left node into stack stack.push(cur); cur = cur.left; } cur = stack.pop(); result.add(cur.val); cur = cur.right; } return result; } }
[Leetcode] Binary Tree Inorder Traversal
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原文地址：http://www.cnblogs.com/andrew-chen/p/4229552.html
