LeetCode-101. Symmetric Tree(Java)

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree 

[1,2,2,3,4,4,3]

 is symmetric:

1
/ \
2   2
/ \ / \
3  4 4  3

But the following 

[1,2,2,null,3,null,3]

 is not:

1
/ \
2   2
\   \
3    3

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题意

判断二叉树是否是镜子二叉树，也就是二叉树以根节点的对称。

思路

最简单的思路就是先从左子树然后右子树遍历，记录遍历结果，然后再右子树左子树遍历，记录遍历结果，然后对比

两个遍历结果，看是否相等。

代码

public class Solution {
public boolean isSymmetric(TreeNode root) {
if(root == null)
return true;
StringBuilder builderOfLeft = new StringBuilder();
StringBuilder builderOfRight = new StringBuilder();
String traverseLeft = traverseLeft(root,builderOfLeft);
String traverseRight = traverseRight(root,builderOfRight);
if(traverseLeft.equals(traverseRight)){
return true;
}
return false;
}
public static String traverseLeft(TreeNode root,StringBuilder builder){
if(root == null){
builder.append("null");
return null;
}
builder.append(root.val+"");
traverseLeft(root.left,builder);
traverseLeft(root.right,builder);
return builder.toString();
}
public static String traverseRight(TreeNode root,StringBuilder builder){
if(root == null){
builder.append("null");
return null;
}
builder.append(root.val+"");
traverseRight(root.right,builder);
traverseRight(root.left,builder);
return builder.toString();
}
}

这个耗时比较长，另外一种实现方式


public class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) {
return true;
}
return isSymmetric(root.left, root.right);
}

public boolean isSymmetric(TreeNode left, TreeNode right) {
if (left == null && right == null) {
return true;
}

if (left == null || right == null) {
return false;
}

return left.val == right.val && isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
}
}