二叉树的前中后序遍历,合并,翻转操作

  • 二叉树的遍历(traversing binary tree)是指从根结点出发,按照某种次序依次访问二叉树中所有的结点,使得每个结点被访问依次且仅被访问一次。
前序
中序
后序


 
java代码实现:

public class TwoTree {
    public static class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode(int x) {
            val = x;
        }
    }
    public static void main(String args[]) {
        TreeNode root = new TreeNode(0); // 构建简单的二叉树
        TreeNode node1 = new TreeNode(1);
        TreeNode node2 = new TreeNode(2);
        TreeNode node3 = new TreeNode(3);
        TreeNode node4 = new TreeNode(4);
        TreeNode node5 = new TreeNode(5);
        TreeNode node6 = new TreeNode(6);
        TreeNode node7 = new TreeNode(7);
        root.left = node1;
        root.right = node2;
        node1.left = node3;
        node1.right = node4;
        node2.left = node5;
        node2.right = node6;
        node4.right = node7;

        preOrderTravels(root);          // 前序遍历一次
        System.out.println();

        middleOrderTravels(root);          // 中序遍历一次
        System.out.println();

        lastOrderTravels(root);          // 后序遍历一次
        System.out.println();

        root = invertBinaryTree(root);
        preOrderTravels(root);          // 翻转后再前序遍历一次

        root = mergeTrees(root, root); //合并两个树
        System.out.println();
        preOrderTravels(root);

    }

    // 前序遍历
    public static void preOrderTravels(TreeNode node) {
        if (node == null) {
            return;
        } else {
            System.out.print(node.val + " ");
            preOrderTravels(node.left);
            preOrderTravels(node.right);
        }
    }

    // 中序遍历
    public static void middleOrderTravels(TreeNode node) {
        if (node == null) {
            return;
        } else {
            middleOrderTravels(node.left);
            System.out.print(node.val + " ");
            middleOrderTravels(node.right);
        }
    }

    // 后序遍历
    public static void lastOrderTravels(TreeNode node) {
        if (node == null) {
            return;
        } else {
            lastOrderTravels(node.left);
            lastOrderTravels(node.right);
            System.out.print(node.val + " ");
        }
    }

    // 翻转二叉树
    private static TreeNode invertBinaryTree(TreeNode root) {
        if (root == null || (root.left == null && root.right == null))
            return root;// 为空,或没有子树,直接返回
        TreeNode tmp = root.right;               // 右子树存入tmp中
        root.right = invertBinaryTree(root.left);// 先处理左子树,然后接到root的右链接
        root.left = invertBinaryTree(tmp);       // 处理tmp中原来的右子树,然后接到root的左链接
        return root;
    }

    // 合并两个二叉树
    public static TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        TreeNode root = null;
        if (t1 == null && t2 == null)
            return root;
        if (t1 == null && t2 != null) {
            root = t2;
        }
        if (t1 != null && t2 == null) {
            root = t1;
        }
        if (t1 != null && t2 != null) {
            root = new TreeNode(t1.val + t2.val);
            root.left = mergeTrees(t1.left, t2.left);
            root.right = mergeTrees(t1.right, t2.right);
        }
        return root;

    }
}
原文地址:https://www.cnblogs.com/cat520/p/9984291.html