144. 二叉樹的前序遍歷
知識點：二叉樹；遞迴；Morris遍歷
題目描述
給你二叉樹的根節點 root ，返回它節點值的 前序 遍歷。
示例
輸入：root = [1,null,2,3]
輸出：[1,2,3]
輸入：root = []
輸出：[]
輸入：root = [1]
輸出：[1]
輸入：root = [1,2]
輸出：[1,2]
輸入：root = [1,null,2]
輸出：[1,2]
解法一：遞迴
/**
* 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;
}
}
時間複雜度；0(N),每個節點恰好被遍歷一次；
空間複雜度；O(N),遞迴過程中棧的開銷；
解法二:迭代法
入棧一定是先右後左，這樣出來才能使先左後右；
壓入根節點；
1.彈出就列印；
2.如有右孩子，壓入右；
3.如有左孩子，壓入左；
/**
* 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遍歷
構建從下到上的連線，一條路能夠走遍所有節點；
/**
* 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){
//有左子樹；
while(mostRightNode.right != null && mostRightNode.right != cur){
mostRightNode = mostRightNode.right; //找到左子樹的最右節點；
}
if(mostRightNode.right == null){
mostRightNode.right = cur; //構建向上的連線；
list.add(cur.val);
cur = cur.left;
continue;
}else{
//第二次到節點，斷開連線
mostRightNode.right = null;
cur = cur.right;
}
}else{
list.add(cur.val);
cur = cur.right;
}
}
return list;
}
}
體會
二叉樹的遍歷是二叉樹的最基礎的，不僅要掌握遞迴寫法，迭代法和morris遍歷也需要掌握，更多詳細可參考這篇總結二叉樹