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?
题意:中序遍历儿二叉树
分类:二叉树
解法1:递归
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- public class Solution {
- public List<Integer> inorderTraversal(TreeNode root) {
- List<Integer> res = new ArrayList<Integer>();
- helper(root,res);
- return res;
- }
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- void helper(TreeNode p,List<Integer> res){
- if(p==null) return;
- if(p.left!=null){
- helper(p.left, res);
- }
- res.add(p.val);
- if(p.right!=null){
- helper(p.right, res);
- }
- }
- }
解法2:非递归
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- public class Solution {
- public List<Integer> inorderTraversal(TreeNode root) {
- Stack<TreeNode> stack = new Stack<TreeNode>();
- List<Integer> res = new ArrayList<Integer>();
- TreeNode cur = root;
- while(stack.size()>0||cur!=null){
- while(cur!=null){
- stack.add(cur);
- cur = cur.left;
- }
- if(stack.size()>0){
- TreeNode p = stack.pop();
- res.add(p.val);
- cur = p.right;
- }
- }
- return res;
- }
- }
原文链接http://blog.youkuaiyun.com/crazy__chen/article/details/46445383