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]
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?
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/binary-tree-inorder-traversal
解法1:
/**
* 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> result = new ArrayList<Integer>();
if(root == null){
return result;
}
if(root.left != null){
result.addAll(inorderTraversal(root.left));
}
result.add(root.val);
if(root.right != null){
result.addAll(inorderTraversal(root.right));
}
return result;
}
}
时间复杂度:O(n) 空间复杂度:O(n)
第二种解法:
/**
* 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> result = new ArrayList<Integer>();
if(root == null){
return result;
}
Stack<TreeNode> stack = new Stack<TreeNode>();
while(root != null || !stack.isEmpty()){
while(root != null){
stack.push(root);
root= root.left;
}
root = stack.pop();
result.add(root.val);
root = root.right;
}
return result;
}
}
时间复杂度:O(n) 空间复杂度O(n)