101.对称二叉树

leetcode链接

代码实现

• 递归

class Solution {public boolean isSymmetric(TreeNode root) {if(root == null) return true;return compare(root.left,root.right);}public boolean compare(TreeNode left,TreeNode right){if(left == null && right == null) return true;else if(left != null && right == null) return false;else if(left == null && right != null) return false;else if(left.val != right.val) return false;boolean out = compare(left.left,right.right);boolean in = compare(left.right,right.left);return out && in;}
}

• 迭代

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 left = queue.poll();TreeNode right = queue.poll();if(left == null && right == null) continue;if(left == null || right == null || left.val != right.val) return false;queue.offer(left.left);queue.offer(right.right);queue.offer(left.right);queue.offer(right.left);}return true;}
}

分析

• 递归法
  • 传入左节点和右节点
  • 判断一下他们的值是否为空，是否相等
  • 再传入外侧的节点，然后传入内侧的节点
  • 内外都对称才能返回true
• 迭代法
  • 使用队列来存储两个节点
  • 每个节点成对加入成对取出比较

相关题目

相同的树

代码

• 递归

class Solution {public boolean isSameTree(TreeNode p, TreeNode q) {if(p == null && q == null) return true;else if(p == null && q != null) return false;else if(p != null && q == null) return false;else if(p.val != q.val) return false;else return isSameTree(p.left,q.left) && isSameTree(p.right,q.right);   }
}

• 迭代

class Solution {public boolean isSameTree(TreeNode p, TreeNode q) {if(p == null && q == null) return true;Queue<TreeNode> queue = new LinkedList<>();queue.offer(p);queue.offer(q);while(!queue.isEmpty()){p = queue.poll();q = queue.poll();if(p == null && q == null) continue;if(p == null || q == null || p.val != q.val) return false;queue.offer(p.left);queue.offer(q.left);queue.offer(p.right);queue.offer(q.right);}return true;}
}

另一棵树的子树

代码

• 双递归

class Solution {public boolean isSubtree(TreeNode root, TreeNode subRoot) {if(root == null && subRoot == null) return true;else if(root == null || subRoot == null) return false;else return isSameTree(root,subRoot) || isSubtree(root.left,subRoot) || isSubtree(root.right,subRoot);}public boolean isSameTree(TreeNode p,TreeNode q){if(p == null && q == null) return true;else if(p == null || q == null) return false;else if(p.val != q.val) return false;else return isSameTree(p.left,q.left) && isSameTree(p.right,q.right);}
}

) return false;
else return isSameTree(p.left,q.left) && isSameTree(p.right,q.right);
}
}