题目 Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree [1,null,2,3],
1

2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?

1,递归

/**
 * 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) {
        List<Integer> result = new ArrayList<Integer>();
        dfs(root,result);
        return result;
    }
    
    private void dfs(TreeNode root, List<Integer> result){
        if(root == null){
            return;
        }
        dfs(root.left,result);
        result.add(root.val);
        dfs(root.right,result);
    }
}

时间复杂度O(n),空间复杂度O(n)

2,使用栈

/**
 * 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) {
        List<Integer> result = new ArrayList<Integer>();
        if(root == null){
           return result;
        }
       
       Stack<TreeNode> stack = new Stack<TreeNode>();
       TreeNode tempNode = root;
       do{
           //左子树不为空,一直入栈
            while (tempNode != null){
                stack.push(tempNode);
                tempNode = tempNode.left;
            }
            //若左子树为空,则弹出栈顶元素且遍历其右子树
            if(!stack.empty()){
                TreeNode visitedNode = stack.pop();
                result.add(visitedNode.val);
                tempNode = visitedNode.right;
            }
        
       }while(!stack.empty()  || tempNode != null);
       return result;
    }
}

时间复杂度O(n),空间复杂度O(n)