LeetCode94 Binary Tree Inorder Traversal

LeetCode94 Binary Tree Inorder Traversal

問題來源LeetCode94

問題描述

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?

問題分析

這道題是很簡單的二叉樹的中序遍歷，最常用的就是遞歸實現，但是這道題後面建議嘗試迭代方案。那麼在這裏就提供兩種方案好了。迭代方案就是使用Stack來實現就可以了。

代碼如下

遞歸代碼

/**
 * 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;
        }
        //遍歷left
        help(list,treeNode.left);
        //根節點
        list.add(treeNode.val);
        help(list,treeNode.right);
    }
}

迭代代碼實現

/**
 * 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;
   }

}

LeetCode學習筆記持續更新

GitHub地址 https://github.com/yanqinghe/leetcode

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