1. 对称二叉树

思路：

1. 在根的值一样接着往下判断
2. 判断左树的左子树的值和右树的右子树的值是否相同
3. 判断左树的右子树的值和右树的左子树的值是否相同

    public boolean isSymmetric(TreeNode root) {if (root == null) {return true;}return isSymmetricChild(root.left,root.right);}private boolean isSymmetricChild(TreeNode leftTree,TreeNode rightTree){if (leftTree == null && rightTree == null) {return true;}if ((leftTree == null && rightTree != null) || (leftTree != null && rightTree == null)) {return false;}if (leftTree.val != rightTree.val) {return false;}return isSymmetricChild(leftTree.left, rightTree.right) &&isSymmetricChild(leftTree.right , rightTree.left);}

2.二叉树的构建及遍历

要求：编一个程序，读入用户输入的一串先序遍历字符串，根据此字符串建立一个二叉树（以指针方式存储）。 例如如下的先序遍历字符串： ABC##DE#G##F### 其中“#”表示的是空格，空格字符代表空树。建立起此二叉树以后，再对二叉树进行中序遍历，输出遍历结果。

import java.util.Scanner;
class TreeNode {public char val;public TreeNode left;public TreeNode right;public TreeNode(char val) {this.val = val;}
}// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {public static int i = 0;public static void main(String[] args) {Scanner in = new Scanner(System.in);while (in.hasNextLine()) {String str = in.nextLine();TreeNode root = createTree(str);inorder(root);}}public static TreeNode createTree(String str) {//1.遍历字符串strTreeNode root = null;if(str.charAt(i) != '#') {//2.根据前序遍历创建二叉树root = new TreeNode(str.charAt(i));i++;root.left = createTree(str);root.right = createTree(str);}else {i++;}//3.返回根节点return root;}public static void inorder(TreeNode root) {if (root == null) {return;}inorder(root.left);System.out.print(root.val + " ");inorder(root.right);}
}

3.二叉树的层序遍历

层序遍历：按照从上到下，从左到右的顺序来访问每一层的节点。
思路：**利用队列的先进先出的思想，先从根节点出发，将其压入队列中，接着判断从队列中弹出来的节点的左右孩子；若该节点的左孩子不为null时，将其压入队列。一直循环，当队列为空时，说明已经把该树遍历完了。

/*** 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;*     }* }*/public List<List<Integer>> levelOrder(TreeNode root) {List<List<Integer>> ret = new ArrayList<>();if (root == null) {return ret;}Queue<TreeNode> queue = new LinkedList<>();queue.offer(root);while (!queue.isEmpty()) {int size = queue.size();List<Integer> tmp = new ArrayList<>();// 层序遍历while (size != 0){TreeNode cur = queue.poll();//System.out.print(cur.val+" ");tmp.add(cur.val);size--;if (cur.left != null){queue.offer(cur.left);}if (cur.right != null) {queue.offer(cur.right);}}ret.add(tmp);}return ret;}

    void levelOrder(TreeNode root) {if (root == null) {return;}Queue<TreeNode> queue = new LinkedList<>();queue.offer(root);while (!queue.isEmpty()) {// 层序遍历TreeNode cur = queue.poll();System.out.print(cur.val+" ");if (cur.left != null){queue.offer(cur.left);}if (cur.right != null) {queue.offer(cur.right);}}}

4.给定一个二叉树, 找到该树中两个指定节点的最近公共祖先

思路：root还是在遍历这棵树，遇到p或者q就返回

1. 如果root节点是p或者是q 那么root是最近祖先
2. 如果pq分布在root的左右两侧，那么root就是最近祖先
3. pq在root的同一侧 遇到的第一个就是公共祖先，即左边不为空，右部为空 或者左空，右不空

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if (root == null){return null;}if (root == p || root == q){return root;}// 递归左树或者右树TreeNode leftTree = lowestCommonAncestor(root.left,p,q);TreeNode rightTree = lowestCommonAncestor(root.right,p,q);if (leftTree != null && rightTree != null) {return root;}else if (leftTree != null) {return leftTree;}else {return rightTree;}}

5. 二叉树创建字符串

思路：

1. 根节点直接拼接
2. 左边为空 && 右空 直接返回
3. 左边不为空，右边为空，加“（”
4. 左边为空，右边不为空 直接加一对括号“（）”

    public String tree2str(TreeNode root) {StringBuilder stringBuilder = new StringBuilder();tree2strChild(root,stringBuilder);return stringBuilder.toString();}private void tree2strChild(TreeNode t, StringBuilder stringBuilder){if (t == null) {return;}stringBuilder.append(t.val);if (t.left != null) {stringBuilder.append("(");tree2strChild(t.left,stringBuilder);stringBuilder.append(")");}else {if (t.right == null) {return;}else {stringBuilder.append("()");}}// 判断右树if (t.right != null) {stringBuilder.append("(");tree2strChild(t.right,stringBuilder);stringBuilder.append(")");}else {return;}}

用栈来存放路径上的节点

    public TreeNode lowestCommonAncestor2(TreeNode root, TreeNode p, TreeNode q) {if(root == null) return null;Stack<TreeNode> stackP = new Stack<>();Stack<TreeNode> stackQ = new Stack<>();getPath(root,p,stackP);getPath(root,q,stackQ);int sizeP = stackP.size();int sizeQ = stackQ.size();if(sizeP > sizeQ) {int size = sizeP - sizeQ;while(size != 0) {stackP.pop();size--;}}else {int size = sizeQ - sizeP;while(size != 0) {stackQ.pop();size--;}}//两个栈当中的元素是一样多while(!stackP.isEmpty() && !stackQ.isEmpty()) {if(stackP.peek() == stackQ.peek()) {return stackP.peek();}else{stackP.pop();stackQ.pop();}}return null;}//判断左树没有这个节点 右树没有这个节点 那么当前的root就不是路径上的节点private boolean getPath(TreeNode root,TreeNode node,Stack<TreeNode> stack) {if(root == null || node == null) {return false;}stack.push(root);if(root == node) {return true;}boolean flg = getPath(root.left,node,stack);if(flg) {return true;}boolean flg2 = getPath(root.right,node,stack);if(flg2) {return true;}stack.pop();return false;}