【leetcode】101.Symmetric Tree【java】再帰と非再帰の2つの方法
1832 ワード
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example: Given binary tree
return its level order traversal as:
For example: Given binary tree
[3,9,20,null,null,15,7]
, 3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
// 1:
/*public class Solution {
public boolean isSymmetric(TreeNode root) {
return root == null || isSymmetricHelp(root.left, root.right);
}
private boolean isSymmetricHelp(TreeNode left, TreeNode right){
if (left == null || right == null){
return left == right;
}
if (left.val != right.val){
return false;
}
return (isSymmetricHelp(left.left, right.right) && isSymmetricHelp(left.right, right.left));
}
}*/
// 2:
public class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null){
return true;
}
Stack stack = new Stack();
stack.push(root.left);
stack.push(root.right);
while (!stack.empty()){
TreeNode p1 = stack.pop();
TreeNode p2 = stack.pop();
if (p1 == null && p2 == null){
continue;
}
if (p1 == null || p2 == null || p1.val != p2.val){
return false;
}
stack.push(p1.left);
stack.push(p2.right);
stack.push(p1.right);
stack.push(p2.left);
}
return true;
}
}