leetCode 94.Binary Tree Inorder Traversal（二叉树中序遍历） 解题思路和方法

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.

思路：二叉树的中序遍历，是典型的递归算法。可是题目中建议非递归实现。所以还是有些思考的。

只是算是基础题。感觉是必须掌握的。

代码例如以下（递归实现）：

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
List<Integer> list = new ArrayList<Integer>();
public List<Integer> inorderTraversal(TreeNode root) {
/**
* 中序遍历，先左子树，再根，最后右子树
*/

if(root == null)
return list;
if(root.left != null){
inorderTraversal(root.left);
}
list.add(root.val);
if(root.right != null){
inorderTraversal(root.right);
}
return list;
}
}

非递归实现：

/**
* 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) {
/**
* 非递归实现中序遍历
* 中序遍历，先左子树，再根。最后右子树
*/

List<Integer> list = new ArrayList<Integer>();

if(root == null)
return list;

TreeNode p = root;
Stack<TreeNode> st = new Stack<>();

while(p != null || !st.isEmpty()){
if(p != null){
st.push(p);
p = p.left;
}else{
p = st.pop();
list.add(p.val);
p = p.right;
}
}
return list;
}
}