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  • leetcode--Binary Tree Inorder Traversal

    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.

    /**
     * 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;
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/averillzheng/p/3553063.html
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