LeetCode

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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?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

   1
  /  2   3
    /
   4
         5

思路：非递归形式，利用栈

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

LeetCode

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原文地址：http://blog.csdn.net/u011345136/article/details/45200273