力扣101.对称二叉树（java)

题目来源

101. 对称二叉树 - 力扣（LeetCode）

代码1(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;*     }* }*/
class Solution {public boolean isSymmetric(TreeNode root) {if(root == null) return true;return isMirror(root.left,root.right);}//判断两是否对称private boolean isMirror(TreeNode left, TreeNode right) {if(left==null && right==null) return true;//只有一个为空if(left==null || right==null) return false;if(left.val != right.val) return false;// 左左与右右 左右与右左return isMirror(left.left,right.right) && isMirror(left.right,right.left);}
}

代码分析

两子树是否对称最小规模子问题就是： 

1. 左子树 与 右子树是否堆成

（对称：左子树与右子树是否都为空，两值是否都为空， 不对称：只有一值为空）

2. 左子树的左子树与 右子树的右子树  是否同时与  左子树的右子树与 右子树的左子树 均为对称

代码2（dfs迭代法）

/*** 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 isSymmetric(TreeNode root) {if(root == null) return true;Stack<TreeNode> stack = new Stack<>();stack.push(root.right);stack.push(root.left);while(!stack.isEmpty()) {TreeNode node1 = stack.pop(); //左子树根结点TreeNode node2 = stack.pop(); //右if(node1==null && node2==null) continue;if(node1==null || node2==null || node1.val!=node2.val) return false;// if(node1==null || node2==null) return false;// if( node1.val!=node2.val) return false;//将......逆向入栈 ： 左左与右右  左右与右左// 入栈就是为了 看两子树是否对称stack.push(node2.left);stack.push(node1.right);stack.push(node2.right);stack.push(node1.left);}return true;}
}

代码分析

这里的dfs，是双向深入的dfs。因为需要两子树进行比较。

代码3（bfs迭代法）

/*** 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 isSymmetric(TreeNode root) {if(root == null) return true;Queue<TreeNode> queue = new LinkedList<>();queue.offer(root.left);queue.offer(root.right);while(!queue.isEmpty()) {TreeNode node1 = queue.poll();TreeNode node2 = queue.poll();if(node1==null && node2==null) continue;if(node1==null || node2==null || node1.val!=node2.val) return false;queue.offer(node1.left);queue.offer(node2.right);queue.offer(node1.right);queue.offer(node2.left);}return true;}
}

代码分析

类似dfs，只是逐层处理。