?
?
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?
confused what {1,#,2,3} means? > read more on how binary tree is serialized on OJ.
源码：
Java版本
算法分析：使用栈。时间复杂度O(n)，空间复杂度O(n)
?
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List
inorderTraversal(TreeNode root) {
List
result=new ArrayList
(); Stack
stack=new Stack
(); TreeNode node=root; do { while(node!=null) { stack.push(node); node=node.left; } if(!stack.isEmpty()) { node=stack.pop(); result.add(node.val); node=node.right; } }while(!stack.isEmpty() || node!=null); return result; } }
?