【LeetCode】144. 二叉树的前序遍历
2024-10-10 09:29:01
144. 二叉树的前序遍历
知识点:二叉树;递归;Morris遍历
题目描述
给你二叉树的根节点 root ,返回它节点值的 前序 遍历。
示例
输入:root = [1,null,2,3]
输出:[1,2,3]
输入:root = []
输出:[]
输入:root = [1]
输出:[1]
输入:root = [1,2]
输出:[1,2]
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> preorderTraversal(TreeNode root) {
if(root == null) return list;
list.add(root.val);
preorderTraversal(root.left);
preorderTraversal(root.right);
return list;
}
}
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
入栈一定是先右后左,这样出来才能使先左后右;
压入根节点;
1.弹出就打印;
2.如有右孩子,压入右;
3.如有左孩子,压入左;
/**
* 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> preorderTraversal(TreeNode root) {
Stack<TreeNode> stack = new Stack<>();
List<Integer> list = new ArrayList<>();
if(root != null) stack.push(root);
while(!stack.isEmpty()){
TreeNode top = stack.pop();
list.add(top.val);
if(top.right != null) stack.push(top.right);
if(top.left != null) stack.push(top.left);
}
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> preorderTraversal(TreeNode root) {
if(root == null) return list;
TreeNode cur = root;
TreeNode mostRightNode = null;
while(cur != null){
mostRightNode = cur.left;
if(mostRightNode != null){
//有左子树;
while(mostRightNode.right != null && mostRightNode.right != cur){
mostRightNode = mostRightNode.right; //找到左子树的最右节点;
}
if(mostRightNode.right == null){
mostRightNode.right = cur; //构建向上的连接;
list.add(cur.val);
cur = cur.left;
continue;
}else{
//第二次到节点,断开连接
mostRightNode.right = null;
cur = cur.right;
}
}else{
list.add(cur.val);
cur = cur.right;
}
}
return list;
}
}
体会
二叉树的遍历是二叉树的最基础的,不仅要掌握递归写法,迭代法和morris遍历也需要掌握,更多详细可参考这篇总结二叉树
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