两道题放一块了,
一个是判断两棵树是否相同:
Given two binary trees, write a function to check if they are equal or not.
Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
一个是判断一棵树是否对称:iven a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1 / \ 2 2 / \ / \ 3 4 4 3
But the following is not:
1 / \ 2 2 \ \ 3 3通过这两道题,可以深刻的体会到树本身递归的特性,用递归求解是非常方便的,代码还非常的简洁:
public boolean isSameTree(TreeNode p, TreeNode q) {
if (p == null && q == null) return true;
else if (p == null || q == null) return false;
if (p.val != q.val) return false;
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}
public boolean isSymmetric(TreeNode root) {
if (root == null) return true;
return dfs(root.left, root.right);
}
private boolean dfs(TreeNode left, TreeNode right) {
if (left == null && right == null) return true;
else if (left == null || right == null) return false;
if (left.val != right.val) return false;
return dfs(left.left, right.right) && dfs(left.right, right.left);
}这种解题实在是太美了。。。

本文介绍两种二叉树的经典问题:如何判断两棵二叉树是否完全相同,以及如何判断一棵二叉树是否对称。通过递归的方法给出简洁优美的解决方案。
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