Leetcode101. Symmetric Tree

Leetcode101. Symmetric Tree

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

Note: Bonus points if you could solve it both recursively and iteratively.

解法一 递归

一棵树是对称的等价于它的左子树和右子树两棵树是对称的，问题就转变为判断两棵树是否对称。

• A 的根节点和 B 的根节点是否相等
• A 的左子树和 B 的右子树是否相等，同时 A 的右子树和左子树是否相等。

class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        //把问题变成判断两棵树是否是对称的
        return isSym(root.left, root.right);
    }
    //判断的是根节点为r1和r2的两棵树是否是对称的
    public boolean isSym(TreeNode r1, TreeNode r2){
        if(r1 == null && r2 == null) return true;
        if(r1 == null || r2 == null) return false;
        //这两棵树是对称需要满足的条件：
        //1.俩根节点相等。 2.树1的左子树和树2的右子树，树2的左子树和树1的右子树都得是对称的
        return r1.val == r2.val && isSym(r1.left, r2.right) && isSym(r1.right, r2.left);
    }
}

DFS 栈

解法一其实就是类似于 DFS 的先序遍历。不同之处是对于 left 子树是正常的先序遍历 根节点 -> 左子树 -> 右子树 的顺序，对于 right 子树的话是 根节点 -> 右子树 -> 左子树 的顺序。所以我们可以用栈，把递归改写为迭代的形式。

public boolean isSymmetric(TreeNode root) { 
    if (root == null) {
        return true;
    }
    Stack<TreeNode> stackLeft = new Stack<>();
    Stack<TreeNode> stackRight = new Stack<>();
    TreeNode curLeft = root.left;
    TreeNode curRight = root.right;
    while (curLeft != null || !stackLeft.isEmpty() || curRight!=null || !stackRight.isEmpty()) {
        // 节点不为空一直压栈
        while (curLeft != null) {
            stackLeft.push(curLeft);
            curLeft = curLeft.left; // 考虑左子树
        }
        while (curRight != null) {
            stackRight.push(curRight);
            curRight = curRight.right; // 考虑右子树
        }
        //长度不同就返回 false
        if (stackLeft.size() != stackRight.size()) {
            return false;
        }
        // 节点为空，就出栈
        curLeft = stackLeft.pop();
        curRight = stackRight.pop();

        // 当前值判断
        if (curLeft.val != curRight.val) {
            return false;
        }
        // 考虑右子树
        curLeft = curLeft.right;
        curRight = curRight.left;
    }
    return true;
}

BFS

一层一层的遍历两个树，然后判断对应的节点是否相等即可。利用两个队列来保存下一次遍历的节点即可。

public boolean isSymmetric6(TreeNode root) {
    if (root == null) {
        return true;
    }
    Queue<TreeNode> leftTree = new LinkedList<>();
    Queue<TreeNode> rightTree = new LinkedList<>();
    //两个树的根节点分别加入
    leftTree.offer(root.left);
    rightTree.offer(root.right);
    while (!leftTree.isEmpty() && !rightTree.isEmpty()) {
        TreeNode curLeft = leftTree.poll();
        TreeNode curRight = rightTree.poll();
        if (curLeft == null && curRight != null || curLeft != null && curRight == null) {
            return false;
        }
        if (curLeft != null && curRight != null) {
            if (curLeft.val != curRight.val) {
                return false;
            }
            //先加入左子树后加入右子树
            leftTree.offer(curLeft.left);
            leftTree.offer(curLeft.right);

            //先加入右子树后加入左子树
            rightTree.offer(curRight.right);
            rightTree.offer(curRight.left);
        }
    }
    if (!leftTree.isEmpty() || !rightTree.isEmpty()) {
        return false;
    }
    return true;
}