【数据结构】:二叉树的非递归前(先)序遍历

给你二叉树的根节点 root ,返回它节点值的 前序 遍历。

示例 1:


输入:root = [1,null,2,3]
输出:[1,2,3]


示例 2:

输入:root = []
输出:[]


示例 3:

输入:root = [1]
输出:[1]


示例 4:


输入:root = [1,2]
输出:[1,2]


示例 5:


输入:root = [1,null,2]
输出:[1,2]

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/binary-tree-preorder-traversal
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

/**
 * 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;
 *     }
 * }
 */
//非递归先序遍历
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root==null){
            return res;
        }
        // 1. 先搞一个栈
        Stack<TreeNode> stack = new Stack<>();
        // 2. 把根节点入栈
        stack.push(root);
        // 3. 循环取栈顶元素并出栈
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            // 访问该元素
            res.add(cur.val);
            // 把右子树入栈
            if (cur.right != null) {
                stack.push(cur.right);
            }
            // 把左子树入栈
            if (cur.left != null) {
                stack.push(cur.left);
            }
        }
        return res;
    }
}

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转载自blog.csdn.net/qq_44292334/article/details/115057227
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