Binary Tree Inorder Traversal

39% Accepted

Given a binary tree, return the inorder traversal of its nodes' values.

Have you met this question in a real interview? Yes
Example
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3


return [1,3,2].

Challenge

Can you do it without recursion?

Tags Expand

• Recursion
• Binary Tree
• Binary Tree Traversal

非递归

/**
 * 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
        if(root == null){
            return new ArrayList<Integer>();
        }

        ArrayList<Integer> result = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();

        TreeNode cur = root;
        while(!stack.isEmpty() || cur != null){
            while(cur != null){
                stack.push(cur);
                cur = cur.left;
            }

            cur = stack.peek();
            result.add(cur.val);
            stack.pop();
            cur = cur.right;
        }

        return result;
    }
}

递归

public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Inorder in ArrayList which contains node values.
     */
    ArrayList<Integer> result = new ArrayList<Integer>();

    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        // write your code here
        if (root == null) {
            return result;
        }

        ArrayList<Integer> left = inorderTraversal(root.left);
        result.add(root.val);
        ArrayList<Integer> right = inorderTraversal(root.right);
        return result;
    }
}