144_二叉树的前序遍历.md

import java.util.*;

/**
 * @author : xfhy
 * Create time : 2020年10月15日09:36:12
 * Description : 144. 二叉树的前序遍历
 * source : https://leetcode-cn.com/problems/binary-tree-preorder-traversal/
 */
public class Solution {

    public static 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;
        }
    }

    //前序遍历: 即先遍历根节点,再遍历左右节点

    //深度优先搜索(DFS) : 采用深度作为优先级,以便从根开始一直到达某个确定的叶子,然后再返回根到达另一个分支.深度优先搜索策略又可以根据根节点,左孩子和右孩子的相对顺序被细分为前序遍历,中序遍历,后续遍历
    //宽度优先搜索(BFS) : 按照高度顺序一层一层的访问整棵树,高层次的节点将会比低层次的节点先被访问到

    /**
     * 递归思路: DFS 就是先遍历根节点,再遍历左右节点,easy~
     */
    public static List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> preList = new LinkedList<>();
        preHelper(root, preList);
        return preList;
    }

    private static void preHelper(TreeNode root, List<Integer> preList) {
        if (root == null) {
            return;
        }
        preList.add(root.val);
        preHelper(root.left, preList);
        preHelper(root.right, preList);
    }

    /**
     * 迭代思路: BFS
     * 从根节点开始,每次迭代弹出栈顶节点,然后压入子节点,先压右节点,再压左节点. 每次弹出栈顶的时候将栈顶节点的数据存入结果List中.
     * 最后的输出顺序是由上往下,由左往右  是符合前序遍历的顺序的
     */
    public static List<Integer> preorderTraversal2(TreeNode root) {
        LinkedList<TreeNode> stack = new LinkedList<>();
        LinkedList<Integer> preList = new LinkedList<>();
        if (root == null) {
            return preList;
        }
        stack.add(root);
        while (!stack.isEmpty()) {
            TreeNode pop = stack.pop();
            preList.add(pop.val);
            if (pop.right != null) {
                stack.push(pop.right);
            }
            if (pop.left != null) {
                stack.push(pop.left);
            }
        }
        return preList;
    }

    //------------------------------------扩展---------------------------------

    /**
     * 迭代算法 BFS  中序遍历(前中后)
     * 从根节点开始找二叉树的最左节点,将走过的节点保存在一个栈中,找到最左节点后访问,对于每个节点来说,它都是以自己为根的子树的根节点,访问完之后就可以转到右儿子上了.
     */
    public static List<Integer> middleOrderTraversal(TreeNode root) {
        LinkedList<TreeNode> stack = new LinkedList<>();
        LinkedList<Integer> preList = new LinkedList<>();
        if (root == null) {
            return preList;
        }
        TreeNode curr = root;
        while (!stack.isEmpty() || curr != null) {
            //找到最左边的节点,依次压入沿线所有节点
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            curr = stack.pop();
            preList.add(curr.val);
            curr = curr.right;
        }
        return preList;
    }

    public static TreeNode createBinaryTree(LinkedList<Integer> inputList) {
        if (inputList == null || inputList.isEmpty()) {
            return null;
        }
        TreeNode node = null;
        Integer data = inputList.removeFirst();
        if (data != null) {
            node = new TreeNode(data);
            node.left = createBinaryTree(inputList);
            node.right = createBinaryTree(inputList);
        }
        return node;
    }

    public static void main(String[] args) {
        /*LinkedList<Integer> integers = new LinkedList<>(Arrays.asList(5, 4, 11, 7, null, null, 2, null, null, null, 8, 13, null, null, 4,
                5, null, null, 1));*/
        /*LinkedList<Integer> integers = new LinkedList<>(Arrays.asList(3, 2, 9, null, null, 10, null,
                null, 8, null, 4));*/
        LinkedList<Integer> integers = new LinkedList<>(Arrays.asList(3, 9, null, null, 20, 15, null, null, 7));
        TreeNode binaryTree = createBinaryTree(integers);

        System.out.println(preorderTraversal(binaryTree));
    }

}