【LeetCode】94. 二叉树的中序遍历
94. 二叉树的中序遍历
知识点：二叉树；递归；Morris遍历
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
输入：root = [1,null,2,3]
输出：[1,3,2]
输入：root = []
输出：[]
输入：root = [1]
输出：[1]
输入：root = [1,2]
输出：[2,1]
输入：root = [1,null,2]
输出：[1,2]
解法一：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 {
List<Integer> list = new ArrayList<>();
public List<Integer> inorderTraversal(TreeNode root) {
if(root == null) return list;
inorderTraversal(root.left);
list.add(root.val);
inorderTraversal(root.right);
return list;
}
}
- python
class Solution:
def inorder(self, root, res):
if not root:
return
self.inorder(root.left, res)
res.append(root.val)
self.inorder(root.right, res)
def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
res = []
self.inorder(root, res)
return res
时间复杂度：0(N),每个节点恰好被遍历一次；
空间复杂度：O(N),递归过程中栈的开销；
解法二:迭代法
1.整条左边界依次入栈；
2.条件1执行不了了，弹出就打印；
3.来到弹出节点的右节点，继续执行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 {
List<Integer> list = new ArrayList<>();
public List<Integer> inorderTraversal(TreeNode root) {
Stack<TreeNode> stack = new Stack<>();
while(!stack.isEmpty() || root != null){
if(root != null){
stack.push(root); //左边一路走到头；
root = root.left;
}else{
TreeNode top = stack.pop();
list.add(top.val);
root = top.right;
}
}
return list;
}
}
解法三：Morris遍历
构建从下到上的连接，一条路能够走遍所有节点；
/**
* 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 {
List<Integer> list = new ArrayList<>();
public List<Integer> inorderTraversal(TreeNode root) {
TreeNode cur = root;
TreeNode mostRightNode = null;
while(cur != null){
mostRightNode = cur.left;
if(mostRightNode != null){
while(mostRightNode.right != null && mostRightNode.right != cur){
mostRightNode = mostRightNode.right;
}
if(mostRightNode.right == null){
mostRightNode.right = cur;
cur = cur.left;
continue;
}else{
mostRightNode.right = null;
list.add(cur.val); //第二次到的时候打印；
cur = cur.right;
}
}else{
list.add(cur.val);
cur = cur.right;
}
}
return list;
}
}
体会
中序遍历的特殊在于其迭代法；
