力扣94. 二叉树的中序遍历 递归&迭代
给定一个二叉树的根节点 root ，返回 它的 中序 遍历 。
示例 1：
输入：root = [1,null,2,3] 输出：[1,3,2]
示例 2：
输入：root = [] 输出：[]
示例 3：
输入：root = [1] 输出：[1]
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 {
public List<Integer> inorderTraversal(TreeNode root) {
if(root == null){
return new ArrayList<>();
}
List<Integer> result = new ArrayList<>();
inTraverse(root,result);
return result;
}
public void inTraverse(TreeNode root,List<Integer> result){
if(root == null){
return;
}
inTraverse(root.left,result);
result.add(root.val);
inTraverse(root.right,result);
}
}
迭代
/**
* 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) {
if (root == null) {
return new ArrayList<>();
}
Stack<TreeNode> stack = new Stack<TreeNode>();
List<Integer> result = new ArrayList<Integer>();
while (stack.size() > 0 || root != null) {
if (root != null) {
stack.push(root);
root = root.left;
} else {
TreeNode temp = stack.pop();
result.add(temp.val);
root = temp.right;
}
}
return result;
}
}
