Symmetric Tree——LeetCode
Given 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
题意就是给定一个二叉树,判定它是否是自己的镜像,这道题我一开始想着只求一下中序序列,然后反着求一下中序序列(右、根、左)两个相等即可。WA了之后,我想了下,某些情况还是不对的,一颗二叉树是自己的镜像,意味着基于根对称,那么求出原来的树的中序、后序,再求出镜像的树的中序、后序,比较一下是否相等即可。当然这种方法不是很好,因为要求四次。
Talk is cheap>>
public boolean isSymmetric(TreeNode root) {
if (root==null)
return true;
return getMidOrderSeq(root).equals(getReMidOrderSeq(root))&&getPostOrderSeq(root).equals(getRePostOrderSeq(root));
} public String getMidOrderSeq(TreeNode node) {
if (node == null) {
return "";
}
return getMidOrderSeq(node.left) + node.val + getMidOrderSeq(node.right);
}
public String getReMidOrderSeq(TreeNode node) {
if (node == null) {
return "";
}
return getReMidOrderSeq(node.right) + node.val + getReMidOrderSeq(node.left);
}
public String getPostOrderSeq(TreeNode node) {
if (node == null) {
return "";
}
return getPostOrderSeq(node.left) +getPostOrderSeq(node.right)+ node.val ;
}
public String getRePostOrderSeq(TreeNode node) {
if (node == null) {
return "";
}
return getRePostOrderSeq(node.right) +getRePostOrderSeq(node.left)+ node.val ;
}
第二种方法就是用递归,方法参数为两个TreeNode,如果一个为null,检查另一个是否为null,否则就检查这两个节点值是否相等,并递归检查这两个节点的左右子树。
public boolean isSymmetric(TreeNode root) {
if (root == null)
return true;
return isSymmetric(root, root); } public boolean isSymmetric(TreeNode left, TreeNode right) {
if (left == null || right == null) {
return left == right;
}
return left.val == right.val && isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
}
第三种方法就是用栈来代替递归,定义两个栈,一左一右,分别将左节点的左孩子(左栈)、右节点的右孩子(右栈)和左节点的右孩子(左栈)、右节点的左孩子(右栈)入栈,循环不变式是两个栈都非空,然后取出栈顶元素,比较是否相等(都为空或值相等),不等则直接return false;然后再将这两个左右节点的左右孩子入栈。。。
public boolean isSymmetricIter(TreeNode root) {
if (root == null || (root.left == null && root.right == null))
return true;
Stack<TreeNode> leftStack = new Stack<>();
Stack<TreeNode> rightStack = new Stack<>();
leftStack.push(root.left);
rightStack.push(root.right);
while (!leftStack.isEmpty() && !rightStack.isEmpty()) {
TreeNode left = leftStack.pop();
TreeNode right = rightStack.pop();
if (left == null && right == null) {
continue;
}
if (left == null || right == null)
return false;
if (left.val != right.val)
return false;
leftStack.push(left.left);
rightStack.push(right.right);
leftStack.push(left.right);
rightStack.push(right.left);
}
return true;
}
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