题目描述
Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree{1,#,2,3},
实例
return[1,3,2].
- 思路
中序遍历二叉树并用链表保留结果返回。
第一种方法是递归；第二种方法是非递归。具体见代码，刚开始非递归没写对！
附加：后序遍历非递归写法
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.ArrayList;
import java.util.Stack;
public class Solution {
public ArrayList<Integer> inorderTraversal(TreeNode root) {
ArrayList<Integer> list = new ArrayList<Integer>();
if(root == null)
return list;
//非递归
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode node = root;
while(stack.isEmpty() == false || node != null){
while(node != null){
stack.push(node);
node = node.left;
}
node = stack.pop();
list.add(node.val);
node = node.right;
}
//递归
//getList(root, list);
return list;
}
public void getList(TreeNode node, ArrayList<Integer> list){
if(node != null){
getList(node.left, list);
list.add(node.val);
getList(node.right, list);
}
}
}
