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);    }}