LeetCode 94. Binary Tree Inorder Traversal（二叉树的中序遍历）

原题网址：https://leetcode.com/problems/binary-tree-inorder-traversal/

Given a binary tree, return the inorder traversal of its nodes' values.

For example:

Given binary tree

{1,#,2,3}

,

1
\
2
/
3

return

[1,3,2]

.

Note: Recursive solution is trivial, could you do it iteratively?

confused what

"{1,#,2,3}"

means? >
read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:

1
/ \
2   3
/
4
\
5

The above binary tree is serialized as

"{1,2,3,#,#,4,#,#,5}"

.
方法一：递归。

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private List<Integer> traversal = new ArrayList<>();
private void traverse(TreeNode root) {
if (root == null) return;
if (root.left != null) traverse(root.left);
traversal.add(root.val);
if (root.right != null) traverse(root.right);
}
public List<Integer> inorderTraversal(TreeNode root) {
traverse(root);
return traversal;
}
}

方法二：将方法一的递归之间修改为用栈实现。

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> traversal = new ArrayList<>();
Stack<TreeNode> nodes = new Stack<>();
Stack<Integer> states = new Stack<>();
nodes.push(root);
states.push(1);
while (!nodes.isEmpty()) {
TreeNode node = nodes.pop();
int state = states.pop();
if (node == null) continue;
if (state == 1) {
nodes.push(node);
states.push(2);
nodes.push(node.left);
states.push(1);
continue;
} else if (state == 2) {
traversal.add(node.val);
nodes.push(node);
states.push(3);
nodes.push(node.right);
states.push(1);
continue;
} else {
// nothing to do
}
}
return traversal;
}
}

方法三：通过prev来区分当前是正在向下遍历，还是已经遍历完成，以免重复压栈。

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> traversal = new ArrayList<>();
if (root == null) return traversal;
Stack<TreeNode> nodes = new Stack<>();
TreeNode prev = null;
nodes.push(root);
while (!nodes.isEmpty()) {
TreeNode node = nodes.pop();
if (prev == null || node == prev.left || node == prev.right) {
if (node.left != null && node.right != null) {
nodes.push(node.right);
nodes.push(node);
nodes.push(node.left);
} else if (node.left != null) {
nodes.push(node);
nodes.push(node.left);
} else if (node.right != null) {
traversal.add(node.val);
nodes.push(node.right);
} else {
traversal.add(node.val);
}
} else {
traversal.add(node.val);
}
prev = node;
}
return traversal;
}
}