leetcode--Binary Tree Inorder Traversal

最新推荐文章于 2024-02-06 03:28:26 发布

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最新推荐文章于 2024-02-06 03:28:26 发布

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原文链接：http://www.cnblogs.com/averillzheng/p/3553063.html

本文介绍了一种不使用递归的二叉树中序遍历算法，通过栈和哈希集来跟踪节点状态，实现了对二叉树节点值的中序遍历。该方法避免了递归可能导致的栈溢出问题，适用于大规模二叉树的遍历。

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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?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        if(root != null){
            Stack<TreeNode> sta = new Stack<TreeNode>();
            sta.push(root);
            HashSet<TreeNode> hset = new HashSet<TreeNode>();
            hset.add(root);
            while(!sta.empty()){
                TreeNode aNode = sta.pop();
                if(aNode.right != null && hset.add(aNode.right)){
                    sta.push(aNode.right);
                    sta.push(aNode);
                    hset.add(aNode.right);
                }
                else if(aNode.left != null && hset.add(aNode.left)){
                    sta.push(aNode);
                    sta.push(aNode.left);
                    hset.add(aNode.right);
                }
                else
                    result.add(aNode.val);                   
            }
        }
        return result;
    }
}

转载于:https://www.cnblogs.com/averillzheng/p/3553063.html