leetcode-java-101. Symmetric Tree

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思路一：
递归：1ms
先判断左右子节点是否相等，不等就返回false
相等则判断左节点的左孩子和右节点的右孩子是否相等，且判断左节点的右孩子和右节点的左孩子是否相等。
思路二：
迭代

public class Solution {     public boolean isSymmetric(TreeNode root) {         if(root == null) {             return true;         } else {             return isSame(root.left,root.right);         }     }     public boolean isSame(TreeNode left,TreeNode right) {         if(left == null && right == null) {             return true;         }         if(left == null || right == null) {             return false;         }         if(left.val != right.val) {             return false;         }         return isSame(left.left,right.right) && isSame(left.right,right.left);     } }

public class Solution {    public boolean isSymmetric(TreeNode root) {        if(root == null) {            return true;        }        Queue<TreeNode> left = new LinkedList<TreeNode>();        Queue<TreeNode> right = new LinkedList<TreeNode>();        if(root.left != null) {            left.offer(root.left);        }        if(root.right != null) {            right.offer(root.right);        }        while(!left.isEmpty() && !right.isEmpty()) {            TreeNode l = left.poll();            TreeNode r = right.poll();            if(r == null && l == null) {                continue;            }            if(r == null || l == null) {                return false;            }            if(l.val != r.val) {                return false;            }            left.offer(l.left);            left.offer(l.right);            right.offer(r.right);            right.offer(r.left);        }        if(left.isEmpty() && right.isEmpty()) {            return true;        } else {            return false;        }    }}

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