您的位置:首页 > 其它

Lintcode 二叉树中序遍历

2015-10-08 21:26 393 查看


二叉树中序遍历

给出一棵二叉树,返回其中序遍历

样例

给出二叉树 
{1,#,2,3}
,
1
\
2
/
3

返回 
[1,3,2]
.
二叉树中序遍历:左子树->根节点->右子树。下图二叉树中序遍历结果为:debgfac。



通过递归方式实现二叉树中序遍历的代码如下:
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: Inorder in ArrayList which contains node values.
*/
ArrayList<Integer> list = new ArrayList<Integer>();
public ArrayList<Integer> inorderTraversal(TreeNode root) {
// write your code here
inorder(root);
return list;

}
private void inorder(TreeNode root){
if(root == null) return;
inorder(root.left);
list.add(root.val);
inorder(root.right);
}

}


通过非递归方式实现二叉树中序遍历的代码如下:
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: Inorder in ArrayList which contains node values.
*/
public ArrayList<Integer> inorderTraversal(TreeNode root) {
// write your code here
Stack<TreeNode> stack = new Stack<TreeNode>();
ArrayList<Integer> list = new ArrayList<Integer>();
while(root!=null||stack.size()>0){
while(root!=null){
stack.push(root);
root = root.left;
}
if(stack.size()>0){
root = stack.pop();
list.add(root.val);
root = root.right;
}

}
return list;
}
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签:  lintcode 二叉树 遍历