-
Difficulty: Medium
-
Related Topics: Hash Table, Stack, Tree
-
Link: https://leetcode.com/problems/binary-tree-inorder-traversal/
Description
Given the root of a binary tree, return the inorder traversal of its nodes' values.
给定一个二叉树的树根 root，返回该二叉树的中序遍历值列表。
Examples
Example 1
Input: root = [1,null,2,3]
Output: [1,3,2]
Example 2
Input: root = []
Output: []
Example 3
Input: root = [1]
Output: [1]
Example 4
Input: root = [1,2]
Output: [2,1]
Example 5
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?
Solution
递归做法是显而易见的，甚至可以直接手撸代码，如下所示：
/**
* Example:
* var ti = TreeNode(5)
* var v = ti.`val`
* Definition for a binary tree node.
* class TreeNode(var `val`: Int) {
* var left: TreeNode? = null
* var right: TreeNode? = null
* }
*/
class Solution {
private val result = arrayListOf<Int>()
fun inorderTraversal(root: TreeNode?): List<Int> {
inOrder(root)
return result
}
private fun inOrder(root: TreeNode?) {
if (root != null) {
inOrder(root.left)
result.add(root.`val`)
inOrder(root.right)
}
}
}
非递归做法需要用到栈来模拟整个过程，代码如下：
/**
* Example:
* var ti = TreeNode(5)
* var v = ti.`val`
* Definition for a binary tree node.
* class TreeNode(var `val`: Int) {
* var left: TreeNode? = null
* var right: TreeNode? = null
* }
*/
import java.util.*
class Solution {
fun inorderTraversal(root: TreeNode?): List<Int> {
if (root == null) {
return emptyList()
}
val result = arrayListOf<Int>()
val stack: Deque<TreeNode> = LinkedList()
var mRoot = root
while (stack.isNotEmpty() || mRoot != null) {
if (mRoot != null) {
// 等价于 inOrder(root.left)
stack.push(mRoot)
mRoot = mRoot.left
} else {
mRoot = stack.pop()
result.add(mRoot.`val`)
// 等价于 inOrder(root.right)
mRoot = mRoot?.right
}
}
return result
}
}