LeetCode

题目：

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?

思路：

用一个栈来存储tree node

package treetraversal;

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

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

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

        return res;
    }

    public static void main(String[] args) {
        // TODO Auto-generated method stub

    }

}