wordpress地址和站点地址区别有没有可以在网站上做试卷的
二叉树的举例
遍历
广度优先遍历
广度优先遍历（Breadth-first-order）：尽可能先访问距离根最近的节点，也称为层序遍历
练习
leetCode102
思路
/*1/ \2 3/ \ / \4 5 6 7头[]尾1 2 3 4 5 6 7
*/实现
public class E01Leetcode102 {public static void main(String[] args) {TreeNode root = new TreeNode(new TreeNode(new TreeNode(4),2,new TreeNode(5)),1,new TreeNode(new TreeNode(6),3,new TreeNode(7)));
System.out.println(testMethod(root));Queue<TreeNode> queue = new LinkedList<>();}public static List<List<TreeNode>> testMethod(TreeNode root){if(root == null){return null;}List<List<TreeNode>> list = new ArrayList<>();LinkedListQueue<TreeNode> queue = new LinkedListQueue();queue.offer(root);int index = 1;while (!queue.isEmpty()){int temp = 0;List<TreeNode> leve = new ArrayList<>();for(int i = 0; i < index; i++){TreeNode n = queue.poll();leve.add(n);if(n.left != null){queue.offer(n.left);temp++;}if(n.right != null){queue.offer(n.right);temp++;}}index = temp;list.add(leve);}return list;}
}深度优先遍历
规则
1.深度优先遍历（Depth-first order）：对于二叉树，可以进一步分成三种（要深入到叶子节点）
前序遍历：先访问该节点，然后是左子树，最后是右子树
中序遍历：先访问左子树，然后是该节点，最后是右子树
后序遍历：先访问左子树，然后是右子树，最后是该节点
实现
前序遍历
LinkedList<TreeNode> list = new LinkedList();
TreeNode curr = root;
while(curr != null || !list.isEmpty){if(curr != null){System.out.println(curr.val);list.push(curr);curr = curr.left;}else{TreeNode pop = list.pop();curr = pop.left}
}中序遍历
LinkedList<TreeNode> list = new LinkedList();
TreeNode curr = root;
while(curr != null || !list.isEmpty){if(curr != null){list.push(curr);curr = curr.left;}else{TreeNode pop = list.pop();curr = pop.leftSystem.out.println(pop.val);}
}后续遍历
static void preOrderEst2(TreeNode node){TreeNode curr = node;LinkedList<TreeNode> list = new LinkedList<>();TreeNode pop = null;while(curr != null || !list.isEmpty()){if(curr != null){list.push(curr);curr = curr.left;}else{TreeNode peek = list.peek();if(peek.right == null || pop == peek.right){pop = list.pop();System.out.println("回" + pop.val);}else {curr = peek.right;}}}}同时解决所有深度优先遍历
static void preOrderEst3(TreeNode node){TreeNode curr = node;LinkedList<TreeNode> list = new LinkedList<>();TreeNode pop = null;while(curr != null || !list.isEmpty()){//待处理的左子树if(curr != null){list.push(curr);System.out.println("前" + curr.val);curr = curr.left;}else{//右子树为空的时候TreeNode peek = list.peek();if(peek.right == null){System.out.println("中" + peek.val);pop = list.pop();System.out.println("后" + pop.val);//右子树处理完毕}else if(pop == peek.right){pop = list.pop();System.out.println("后" + pop.val);// 待处理的右子树}else{System.out.println("中" + peek.val);curr = peek.right;}}}}