94. 二叉树的中序遍历 (opens new window)

English Version (opens new window)

题目描述

给定一个二叉树的根节点 root ，返回它的 中序 遍历。

示例 1：

输入：root = [1,null,2,3]
输出：[1,3,2]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [1]
输出：[1]

示例 4：

输入：root = [1,2]
输出：[2,1]

示例 5：

输入：root = [1,null,2]
输出：[1,2]

提示：

• 树中节点数目在范围 [0, 100] 内
• -100 <= Node.val <= 100

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

解法

递归遍历或利用栈实现非递归遍历。

非递归的思路如下：

1. 定义一个栈
2. 将树的左节点依次入栈
3. 左节点为空时，弹出栈顶元素并处理
4. 重复 2-3 的操作

Python3

递归：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        def inorder(root):
            if root:
                inorder(root.left)
                res.append(root.val)
                inorder(root.right)
        res = []
        inorder(root)
        return res

非递归：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        s = []
        res = []
        while root or s:
            if root:
                s.append(root)
                root = root.left
            else:
                root = s.pop()
                res.append(root.val)
                root = root.right
        return res

Java

递归：

/**
 * 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 {

    private List<Integer> res;

    public List<Integer> inorderTraversal(TreeNode root) {
        res = new ArrayList<>();
        inorder(root);
        return res;
    }

    private void inorder(TreeNode root) {
        if (root != null) {
            inorder(root.left);
            res.add(root.val);
            inorder(root.right);
        }
    }
}

非递归：

/**
 * 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) {
        if (root == null) {
            return Collections.emptyList();
        }
        List<Integer> res = new ArrayList<>();
        Deque<TreeNode> s = new ArrayDeque<>();
        while (root != null || !s.isEmpty()) {
            if (root != null) {
                s.push(root);
                root = root.left;
            } else {
                root = s.pop();
                res.add(root.val);
                root = root.right;
            }
        }
        return res;
    }
}

JavaScript

递归：

var inorderTraversal = function(root) {
    let res = [];
    function inorder(root){
        if(root){
        inorder(root.left);
        res.push(root.val);
        inorder(root.right);
        }
    }
    inorder(root);
    return res;
};

非递归：

var inorderTraversal = function (root) {
    let res = [], stk = [];
    let cur = root;
    while (cur || stk.length !== 0) {
        while (cur) {
            stk.push(cur);
            cur = cur.left;
        } 
        let top = stk.pop();
        res.push(top.val);
        cur = top.right;

    }
    return res;
};

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