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# 94.Binary-Tree-Inorder-Traversal

## 题目描述

Given a binary tree, return the inorder traversal of its nodes' values.
Example:
Input: [1,null,2,3]
1
\
2
/
3
Output: [1,3,2]
Follow up: Recursive solution is trivial, could you do it iteratively?

## 代码

### Approach #1 Recursive Approach

Complexity Analysis
• Time complexity : O(n)
• Space complexity : The worst case space required is O(n), and in the average case it's O(logn) where n is number of nodes.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
dfs(root, res);
return res;
}
public void dfs(TreeNode root, List<Integer> res) {
if (root != null) {
if (root.left != null) {
dfs(root.left, res);
}
if (root.right != null) {
dfs(root.right, res);
}
}
}
}

### Approach #2 Iterating method using stack

Complexity Analysis
• Time complexity : O(n)
• Space complexity : O(n)
1. Push the left node first
2. Pop and visit the value
3. Try to move to the right node
class Solution {
pulic List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
Stack<TreeNode> stack = new Stack<>();
TreeNode curr = root;
while (curr != null || !stack.isEmpty()) {
// 1. Push the left node first
while (curr != null) {
stack.push(curr);
curr = curr.left;
}
// 2. Pop and visit the value
curr = stack.pop();
// 3. Try to move to the right node
curr = curr.right;
}
return res;
}
}

### Approach #3 Morris Traversal

Step 1: Initialize current as root
Step 2: While current is not NULL,
Complexity Analysis
• Time complexity : O(n)
• Space complexity : O(n)
If current does not have left child
b. Go to the right, i.e., current = current.right
Else
a. In current's left subtree, make current the right child of the rightmost node
b. Go to this left child, i.e., current = current.left
class Solution {
public List<Integer> inorder inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
TreeNode curr = root;
TreeNode pre;
while (curr != null) {
if (curr.left == null) {
curr = curr.right;
} else {
pre = curr.left;
while (pre.right != null) {
pre = pre.right;
}
pre.right = curr;
TreeNode temp = curr;
curr = curr.left;
temp.left = null;
}
}
return res;
}
}