LeetCode刷题笔录 Binary Tree Inorder Traversal_given a binary tree, implement the in-order traver-优快云博客

本文介绍了一种二叉树中序遍历的迭代算法实现，该方法通过使用两个栈分别存储节点和计数，实现了无需递归即可完成中序遍历的过程。

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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?

这题和Preorder traversal的区别重点在于一个node第一次被从stack中pop出来时，不对其进行访问，而是要将其放回stack。在第二次pop时再进行访问。

在纸上画一画就可以得到以下要点：

1.需要记录一个node是第几次被从stack中pop

2.第一次pop出一个node后，按照right child, node, left child的顺序将这三个node（如果存在）放回stack

在这里我用了一个Stack<Integer> counterStack来记录每个node是第几次被pop出stack。初始值是0，第一次被pop后数值变为1. 两个stack的元素一一对应，这样就可以知道当前node的count。

public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        if(root == null)
            return result;
        Stack<TreeNode> nodeStack = new Stack<TreeNode>();
        Stack<Integer> counterStack = new Stack<Integer>();
        nodeStack.push(root);
        counterStack.push(0);
        
        while(!nodeStack.isEmpty()){
            TreeNode node = nodeStack.pop();
            int count = counterStack.pop();
            //if this is the second time that this node is popped out, visit it
            if(count == 1){
                result.add(node.val);
            }
            //else push the node's right child to the stack, itself back to the stack, add its counter, and adds its left child
            else{
                if(node.right != null){
                    nodeStack.push(node.right);
                    counterStack.push(0);
                }
                nodeStack.push(node);
                counterStack.push(1);
                if(node.left != null){
                    nodeStack.push(node.left);
                    counterStack.push(0);
                }
            }
        }
        
        return result;
    }
}