leetcode 63: Binary Tree Inorder Traversal

二叉树中序遍历

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原文链接：http://www.cnblogs.com/xishibean/archive/2013/01/29/2951324.html

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#java

本文介绍了一种迭代方式实现二叉树的中序遍历，并提供了详细的Java代码实现。该方法使用栈来辅助遍历过程，避免了递归带来的堆栈溢出风险。

Binary Tree Inorder Traversal Aug 27 '12

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?

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

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

转载于:https://www.cnblogs.com/xishibean/archive/2013/01/29/2951324.html