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Leetcode 94 Binary Tree Inorder Traversal

发表于 更新于 分类于 Valine：
本文字数： 1.9k 阅读时长 ≈ 2 分钟

Given the root of a binary tree, return the inorder traversal of its nodes’ values.

Example 1:

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Input: root = [1,null,2,3]
Output: [1,3,2]

Example 2:

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Input: root = []
Output: []

Example 3:

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Input: root = [1]
Output: [1]

Example 4:

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Input: root = [1,2]
Output: [2,1]

Example 5:

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Input: root = [1,null,2]
Output: [1,2]

Constraints:

  • The number of nodes in the tree is in the range [0, 100].
  • -100 <= Node.val <= 100

Follow up:

Recursive solution is trivial, could you do it iteratively?

1 递归

记得之前看到过这个非递归解法, 但是实在想不出来了😭😭

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        inorderTraversal(ans, root);
        return ans;
    }
    private void inorderTraversal(List<Integer> ans, TreeNode root)
    {
        if(root == null)
            return;
        else
        {
            inorderTraversal(ans, root.left);
            ans.add(root.val);
            inorderTraversal(ans, root.right);
        }
    }
}

2 非递归

具体方法还是看代码吧. 感觉很难描述.

当初看解析的时候也看了半天才懂

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        Stack<TreeNode> stk = new Stack<>();
        List<Integer> inorder = new ArrayList<>();
        TreeNode p = root;
        while(p != null || !stk.isEmpty())
        {
            while(p != null)
            {
                stk.push(p);
                p = p.left;
            }
            
            p = stk.pop();
            inorder.add(p.val);
            p = p.right;
        }
        return inorder;
    }
}