[剑指Offer]-之字形打印二叉树

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题目描述

请实现一个函数按照之字形顺序打印二叉树,即第一行按照从左到右的顺序打印,第二层按照从右到左的顺序打印,第三行再按照从左到右的顺序打印,其他行以此类推。

解题思路

• 使用currentlayer记录层数，从一开始计数
• 当两个栈任意一个不为空时进行循环
• 弹出当前层栈尾元素，同时层数+1，保存分奇偶保存下一层元素
  
• 最后需要注意的时打印换行的时机，分奇偶讨论，如果栈空则回退一层。
  

算法图解

• 注意代码中对于结束递归的条件的设定！
• 是否访问数组的赋值！

参考代码：

package offer;import java.util.Stack;/*** 之形打印二叉树*/
public class Offer32_3 {public static void main(String[] args) {BinaryTreeNode root = new BinaryTreeNode(1);BinaryTreeNode node1 = new BinaryTreeNode(2);BinaryTreeNode node2 = new BinaryTreeNode(3);BinaryTreeNode node3 = new BinaryTreeNode(4);BinaryTreeNode node4 = new BinaryTreeNode(5);BinaryTreeNode node5 = new BinaryTreeNode(6);BinaryTreeNode node6 = new BinaryTreeNode(7);BinaryTreeNode node7 = new BinaryTreeNode(8);BinaryTreeNode node8 = new BinaryTreeNode(9);BinaryTreeNode node9 = new BinaryTreeNode(10);BinaryTreeNode node10 = new BinaryTreeNode(11);BinaryTreeNode node11 = new BinaryTreeNode(12);BinaryTreeNode node12 = new BinaryTreeNode(13);BinaryTreeNode node13 = new BinaryTreeNode(14);BinaryTreeNode node14 = new BinaryTreeNode(15);root.leftTree = node1;root.rightTree = node2;node1.leftTree = node3;node3.leftTree = node7;node3.rightTree = node8;node1.rightTree = node4;node4.leftTree = node9;node4.rightTree = node10;node2.rightTree = node6;node6.leftTree = node13;node6.rightTree = node14;node2.leftTree = node5;node5.leftTree = node11;node5.rightTree = node12;printByZ(root);}private static void printByZ(BinaryTreeNode root) {if (root == null) {return;}int currentlayer = 1;Stack<BinaryTreeNode> stackj = new Stack<BinaryTreeNode>();Stack<BinaryTreeNode> stacko = new Stack<BinaryTreeNode>();stackj.push(root);while (!stackj.empty() || !stacko.empty()) {BinaryTreeNode currNode = null;if (currentlayer % 2 == 0 && !stacko.empty()) {currNode = stacko.pop();}if (currentlayer % 2 != 0 && !stackj.empty()) {currNode = stackj.pop();}System.out.print(currNode.node_value + " ");currentlayer++;if (currentlayer % 2 == 0) {if (currNode.leftTree != null) {stacko.push(currNode.leftTree);}if (currNode.rightTree != null) {stacko.push(currNode.rightTree);}} else {if (currNode.rightTree != null) {stackj.push(currNode.rightTree);}if (currNode.leftTree != null) {stackj.push(currNode.leftTree);}}/*** 分奇偶 讨论什么时候打印换行  放在一起判断的话会出现某一个空就打印换行 引起空指针异常*/if (currentlayer % 2 != 0) {if (stacko.empty()) {System.out.println();} else {currentlayer = currentlayer - 1;}} else {if (stackj.empty()) {System.out.println();} else {currentlayer = currentlayer - 1;}}}}
}

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附录

该题源码在我的 ?Github 上面！

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