94. Binary Tree Inorder Traversal
Given the root of a binary tree, return the inorder traversal of its nodes' values.
Example 1:
Input: root = [1,null,2,3]
Output: [1,3,2]
Example 2:
Input: root = []
Output: []
Example 3:
Input: root = [1]
Output: [1]
Example 4:
Input: root = [1,2]
Output: [2,1]
Example 5:
Input: root = [1,null,2]
Output: [1,2]
Constraints:
- The number of nodes in the tree is in the range [0, 100].
- -100 <= Node.val <= 100
Follow up:
- Recursive solution is trivial, could you do it iteratively?
递归操作1：帮助理解dfs
class Solution {
List<Integer>res=new LinkedList<Integer>();
private void dfs(TreeNode root){
if(root!=null){
dfs(root.left);
res.add(root.val);
dfs(root.right);
}
}
public List<Integer> inorderTraversal(TreeNode root) {
dfs(root);
return res;
}
}递归操作2:
/**
* 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> result = new ArrayList<>();
public List<Integer> inorderTraversal(TreeNode root) {
if(root!=null){
inorderTraversal(root.left);
result.add(root.val);
inorderTraversal(root.right);
}
return result;
}
}非递归操作
/**
* 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> list = new ArrayList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
while(root!=null || !stack.empty()){
while(root!=null){
stack.push(root);
root = root.left;
}
root = stack.pop();
list.add(root.val);
root = root.right;
}
return list;
}
}