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;
}