力扣94.二叉树的中序遍历（递归and迭代法）（java）

题目来源

递归法

java 复制代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        inorder(root,res);
        return res;
    }

    private void inorder(TreeNode root, List<Integer> res) {
        if(root == null) return ;
        // 左中右，这样子记录
        inorder(root.left, res);
        res.add(root.val);
        inorder(root.right, res);
    }
}

迭代法

代码分析

递归啥的不赘述了。迭代法就是模拟递归栈。

因为如果想要实现左中右的遍历效果（也就是中序遍历效果），就需要先找到最左边的，

但是找到最左边怎么回去呢？如果我在沿途找的时候，把过程中的结点存放起来,而栈最合适。

遍历结果就存放在动态数组中，（ArrayList)

代码

java 复制代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        TreeNode curr = root;

        while(curr != null || !stack.isEmpty()) {
            //递归找最左边，入栈
            while(curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            // 回退， 出栈
            curr = stack.pop();
            res.add(curr.val);
            //遍历右边
            curr = curr.right;
        }
        return res;
    }
}