[LeetCode]Binary Tree Inorder Traversal

标签：java   leetcode   

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.

   1
  /  2   3
    /
   4
         5

递归方法

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
    	if(root == null){
    		List<Integer> list = new ArrayList<Integer>();
    		return list;
    	}
    	if(root.left==null&&root.right==null){
    		List<Integer> list = new ArrayList<Integer>();
    		list.add(root.val);
    		return list;
    	}
    	
    	List<Integer> leftList = inorderTraversal(root.left);
    	List<Integer> rightList = inorderTraversal(root.right);
    	List<Integer> res = new ArrayList<>();
    	res.addAll(leftList);
    	res.add(root.val);
    	res.addAll(rightList);
    	return res;
    }
}

stack维护

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

[LeetCode]Binary Tree Inorder Traversal

标签：java   leetcode   

原文地址：http://blog.csdn.net/guorudi/article/details/41315291