Leetcode: Binary Tree Inorder Traversal

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本文介绍了一种实现二叉树中序遍历的方法，包括递归和迭代两种方式。递归方法直接通过调用自身来遍历左子树、访问节点、遍历右子树；迭代方法则使用栈来辅助完成，先将所有左子节点压入栈中，直至遇到叶子节点，然后弹出并访问节点，再将右子节点压入栈，直至遍历完整棵树。

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

For example:
Given binary tree {1,#,2,3},

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
		// Start typing your Java solution below
		// DO NOT write main() function
		ArrayList<Integer> res = new ArrayList<Integer>();
		if(root != null)
			inorder(res, root);
		return res;
	}
	
	public void inorder(ArrayList<Integer> res, TreeNode root){
		if(root.left != null)
			inorder(res, root.left);
		res.add(root.val);
		if(root.right != null)
			inorder(res,root.right);
	}
}

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
		// Start typing your Java solution below
		// DO NOT write main() function
		ArrayList<Integer> res = new ArrayList<Integer>();
		if(root == null)
			return res;
		Stack<TreeNode> stack = new Stack<TreeNode>();
		TreeNode p = root;
		do{
			if(p != null){
				stack.push(p);
				p = p.left;
			}
			else{
				p = stack.pop();
				res.add(p.val);
				p = p.right;
			}
		}while(!stack.empty() || p != null);
		return res;
	}
}