leetcode94. Binary Tree Inorder Traversal

94.Binary Tree Inorder Traversal

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

For example:
Given binary tree [1,null,2,3],
   1
    \
     2
    /
   3
return [1,3,2].

解法一

采用分治策略，递归计算左边节点的中序遍历，右边节点的中序遍历。

/**
 * Definition for a binary tree node.
 * 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> result = new ArrayList<Integer>();
        if (root == null) {
            return result;
        }

        List<Integer> left = inorderTraversal(root.left);
        List<Integer> right = inorderTraversal(root.right);

        result.addAll(left);
        result.add(root.val);
        result.addAll(right);
        return result;
    }
}

解法二

非递归方法，先把左孩子放栈，然后一个一个pop，先添加本节点，再判断有没有右孩子，有得话入栈，再pop。

/**
 * Definition for a binary tree node.
 * 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> result = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode cur = root;
        while (!stack.empty() || cur != null) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            result.add(cur.val);
            cur = cur.right;
        }
        return result;
    }
}