【LeetCode】94.Binary Tree Inorder Traversal(Medium)解题报告

【LeetCode】94.Binary Tree Inorder Traversal(Medium)解题报告

题目地址：https://leetcode.com/problems/binary-tree-level-order-traversal-ii/description/
题目描述：

  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].

Note: Recursive solution is trivial, could you do it iteratively?

Solution1:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 二叉树的中序遍历：左根右
 time : O(n)
 space : O(n)
 递归形式和
 */
class Solution {
    public List inorderTraversal(TreeNode root) {
        List res = new ArrayList<>();
        if(root == null) return res;
        helper(res,root);
        return res;
    }
    public static void helper(List res,TreeNode root){
        if(root == null) return;
        helper(res,root.left);
        res.add(root.val);
        helper(res,root.right);
    }
}

Solution2:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 二叉树的中序遍历：左根右
 time : O(n)
 space : O(n)
 递归形式和迭代stack
 */
class Solution {
    public List inorderTraversal(TreeNode root) {
        List res = new ArrayList<>();
        if(root == null) return res;
        Stack stack = new Stack<>();
        TreeNode cur = root;
        while(cur != null || !stack.isEmpty()){
            while(cur != null){
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            res.add(cur.val);
            cur = cur.right;
        }
        return res;
    }

}

Date：2018年3月17日