LeetCode94 Binary Tree Inorder Traversal

问题来源LeetCode94

问题描述

Given a binary tree, return the inorder traversal of its nodes’ values.
For example:
Given binary tree [1,null,2,3],
1
\
2
/
3

return [1,3,2].
Recursive solution is trivial, could you do it iteratively?

问题分析

这道题是很简单的二叉树的中序遍历，最常用的就是递归实现，但是这道题后面建议尝试迭代方案。那么在这里就提供两种方案好了。迭代方案就是使用Stack来实现就可以了。

代码如下

递归代码

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */class Solution {    public List<Integer> inorderTraversal(TreeNode root) {        List<Integer> list =new ArrayList<>();        help(list,root);        return list;    }    private void help(List<Integer> list,TreeNode treeNode){        if(treeNode==null){            return;        }        //遍历left        help(list,treeNode.left);        //根节点        list.add(treeNode.val);        help(list,treeNode.right);    }}

迭代代码实现

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */class Solution {    public List<Integer> inorderTraversal(TreeNode root) {        List<Integer> result = new LinkedList<>();        TreeNode curr = root, prev = null;        while(curr != null) {            if(curr.left == null) {                result.add(curr.val);                curr = curr.right;            } else {                prev = curr.left;                while(prev.right != null && prev.right != curr)                    prev = prev.right;                if(prev.right == null) {                    prev.right = curr;                    curr = curr.left;                } else {                    result.add(curr.val);                    prev.right = null;                    curr = curr.right;                }            }        }        return result;   }}

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