101. 对称二叉树
力扣题目链接(opens new window)
给定一个二叉树，检查它是否是镜像对称的。
递归
/*** 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;* }* }*/
//递归
class Solution {public boolean compare(TreeNode left,TreeNode right){if(left == null && right == null) return true;//左右节点都为空else if(left != null && right == null) return false;//左不空，右空，不可能对称else if(left ==null && right != null) return false;//右不空，左空，不可能对称else if(left.val != right.val) return false;//左右都不为空，但不相等else{//左右都不为空，且相当，进入下一次的递归判断boolean out = compare(left.left,right.right);//左，右边树的外侧节点判断boolean inside = compare(left.right,right.left);//左，右边树的内侧节点判断return (out&& inside);}}public boolean isSymmetric(TreeNode root) {return compare(root.left,root.right);}
}迭代
/*** 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;* }* }*///迭代
class Solution {public boolean isSymmetric(TreeNode root) {Stack<TreeNode> stack = new Stack<>();//栈辅助if(root == null) return true;stack.push(root.left);stack.push(root.right);while(!stack.isEmpty()){TreeNode left = stack.peek();//获得栈顶结点stack.pop();//弹出栈顶结点TreeNode right = stack.peek();//再次获得栈顶结点stack.pop();//再弹出栈顶结点if(left ==null&&right==null) continue;//左右都为空，继续下一次循环else if(left != null && right == null) return false;else if(left ==null && right != null) return false;else if(left.val != right.val) return false;else{//左右都不为空，且相等stack.push(left.left);stack.push(right.right);stack.push(left.right);stack.push(right.left);}}return true;}
}2，力扣100，相同的树
给你两棵二叉树的根节点 p 和 q ，编写一个函数来检验这两棵树是否相同。如果两个树在结构上相同，并且节点具有相同的值，则认为它们是相同的。示例 1：输入：p = [1,2,3], q = [1,2,3]
输出：true
示例 2：输入：p = [1,2], q = [1,null,2]
输出：false
示例 3：输入：p = [1,2,1], q = [1,1,2]
输出：false提示：两棵树上的节点数目都在范围 [0, 100] 内
-104 <= Node.val <= 104/*** 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;* }* }*/
class Solution {public boolean isSameTree(TreeNode p, TreeNode q) {if(p == null && q == null) return true;//左右节点都为空else if(p != null && q == null) return false;//左不空，右空，不可能对称else if(p ==null && q != null) return false;//右不空，左空，不可能对称else if(p.val != q.val) return false;//左右都不为空，但不相等else{//左右都不为空，且相当，进入下一次的递归判断boolean out = isSameTree(p.left,q.left);boolean inside = isSameTree(p.right,q.right);return (out&& inside);}}
}3 力扣572，另一棵子树
给你两棵二叉树 root 和 subRoot 。检验 root 中是否包含和 subRoot 具有相同结构和节点值的子树。如果存在，返回 true ；否则，返回 false 。
二叉树 tree 的一棵子树包括 tree 的某个节点和这个节点的所有后代节点。tree 也可以看做它自身的一棵子树。
示例 1：
输入：root = [3,4,5,1,2], subRoot = [4,1,2] 输出：true
示例 2：
输入：root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2] 输出：false
/*** 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;* }* }*/
class Solution {public boolean isSametree(TreeNode root,TreeNode subRoot){if(root == null && subRoot == null) return true;//左右节点都为空else if(root != null && subRoot == null) return false;//左不空，右空，不可能对称else if(root ==null && subRoot != null) return false;//右不空，左空，不可能对称else if(root.val != subRoot.val) return false;//左右都不为空，但不相等else{//左右都不为空，且相当，进入下一次的递归判断boolean out = isSametree(root.left,subRoot.left);boolean inside = isSametree(root.right,subRoot.right);return (out&& inside);}}public boolean isSubtree(TreeNode root, TreeNode subRoot) {if(root==null&&subRoot==null) return true;else if(root==null && subRoot!=null) return false;else if(root!=null && subRoot == null) return true;else if(isSametree(root,subRoot)) return true;else{return isSubtree(root.left,subRoot)||isSubtree(root.right,subRoot);}}}
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