94. 二叉樹的中序遍歷
知識點：二叉樹；遞迴；Morris遍歷
題目描述
給定一個二叉樹的根節點 root ，返回它的 中序 遍歷。
示例
輸入：root = [1,null,2,3]
輸出：[1,3,2]
輸入：root = []
輸出：[]
輸入：root = [1]
輸出：[1]
輸入：root = [1,2]
輸出：[2,1]
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> inorderTraversal(TreeNode root) {
if(root == null) return list;
inorderTraversal(root.left);
list.add(root.val);
inorderTraversal(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
1.整條左邊界依次入棧；
2.條件1執行不了了，彈出就列印；
3.來到彈出節點的右節點，繼續執行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); //左邊一路走到頭；
root = root.left;
}else{
TreeNode top = stack.pop();
list.add(top.val);
root = top.right;
}
}
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> 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); //第二次到的時候列印；
cur = cur.right;
}
}else{
list.add(cur.val);
cur = cur.right;
}
}
return list;
}
}
體會
中序遍歷的特殊在於其迭代法；