[LeetCode94]Binary Tree Inorder Traversal

Binary Tree Inorder Traversal

Total Accepted: 80048 Total
Submissions: 219796My Submissions

Question
Solution

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.

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使用递归的方法
Code:

package com.leetcode.problems;

import java.util.ArrayList;
import java.util.List;

class TreeNode {
     int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}
public class Solution94 {

	public void recursion(TreeNode root,List<Integer> re){
		if(root!=null){
			
			if(root.left!=null){
				
				recursion(root.left, re);
			}
			re.add(root.val);
			if(root.right!=null){
				recursion(root.right, re);
			}
		}
	}
	public List<Integer> inorderTraversal(TreeNode root) {
		List<Integer> list = new ArrayList<Integer>();
		if(root != null){
			
			recursion(root,list);
		}
		return list;
		
	}
	public static void main(String[] args) {
		
		TreeNode tn1 = new TreeNode(1);
		
		TreeNode tn2 = new TreeNode(2);
		TreeNode tn3 = new TreeNode(3);
		tn1.left = null;
		tn1.right = tn2;
		tn2.left = tn3;
		tn2.right = null;
		tn3.left = null;
		tn3.right = null;
		List<Integer> list = new Solution94().inorderTraversal(tn1);
		for(Integer i:list){
			System.out.println(i);
		}
	}
}

方式二：使用栈的方式

public List<Integer> inorderTraversal(TreeNode root) {
		List<Integer> list = new ArrayList<Integer>();
		
		Stack<TreeNode> stack = new Stack<TreeNode>();
		TreeNode cur = root;
		while(cur!=null || !stack.isEmpty()){
			while(cur!=null){
				stack.push(cur);
				cur = cur.left;
			}
			cur = stack.pop();
			list.add(cur.val);
			cur = cur.right;
		}
		return list;
		
	}