Binary Tree Inorder Traversal

https://oj.leetcode.com/problems/binary-tree-inorder-traversal/

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?

解题思路：

递归

/**
 * 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) {
        List list = new ArrayList<Integer>();
        if(root == null){
            return list;
        }
        list.addAll(inorderTraversal(root.left));
        list.add(root.val);
        list.addAll(inorderTraversal(root.right));
        return list;
    }
}

非递归方法

/**
 * 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) {
        List<Integer> list = new ArrayList<Integer> ();
        Stack<TreeNode> stack = new Stack<TreeNode> ();
        
        while(!stack.empty() || root != null){
            while(root != null){
                stack.push(root);
                root = root.left;
            }
            if(!stack.empty()){
                root = stack.pop();
                list.add(root.val);
                root = root.right;
            }
        }
        return list;
    }
}

 上面的内层while是redundant的。想想为什么？这里的思路是，将当前节点推进栈，然后将当前节点赋值为它的左节点。如此往复直到左节点为空，这时从栈顶弹出一个元素，处理它。然后再把当前节点的右节点赋值为当前节点。循环上面的过程。

/**
 * 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) {
        List<Integer> list = new ArrayList<Integer> ();
        Stack<TreeNode> stack = new Stack<TreeNode> ();
        
        while(!stack.empty() || root != null){
            if(root != null){
                stack.push(root);
                root = root.left;
            }else if(!stack.empty()){
                root = stack.pop();
                list.add(root.val);
                root = root.right;
            }
        }
        return list;
    }
}