對稱二叉樹

題目

給定一個二叉樹，檢查它是不是鏡像對稱的。

例如，二叉樹[1,2,2,3,4,4,3]是對稱的。

1
   / \
  2   2
 / \ / \
3  4 4  3

可是下面這個[1,2,2,null,3,null,3]則不是鏡像對稱的:

1
   / \
  2   2
   \   \
   3    3

解法

解法一： 遞歸。

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { // 方法一：遞歸 public boolean isSymmetric(TreeNode root) { if(root == null){ return true; } TreeNode left = root.left; TreeNode right = root.right; return isSymmetricForLeftAndRight(left,right); } // 判斷left和right是否對稱 public boolean isSymmetricForLeftAndRight(TreeNode left,TreeNode right){ if(left == null || right == null){ if(left != null || right != null){ return false; } else { return true; } } // 判斷值是否相等 int leftVal = left.val; int rightVal = right.val; if(leftVal != rightVal){ return false; } // 遞歸調用，判斷左節點的左節點與右節點的右節點是否對稱 TreeNode leftNextLeft = left.left; TreeNode rightNextRight = right.right; if(!isSymmetricForLeftAndRight(leftNextLeft,rightNextRight)){ return false; } // 遞歸調用，判斷左節點的右節點與右節點的左節點是否對稱 TreeNode leftNextRight = left.right; TreeNode rightNextLeft = right.left; if(!isSymmetricForLeftAndRight(leftNextRight,rightNextLeft)){ return false; } return true; } } 

解法二： 迭代。

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { // 解法二：迭代 public boolean isSymmetric(TreeNode root) { if(root == null){ return true; } // 建立隊列，特色是先進先出。 Queue<TreeNode> queue = new LinkedList<>(); queue.offer(root.left); queue.offer(root.right); while(!queue.isEmpty()){ TreeNode left = queue.poll(); TreeNode right = queue.poll(); if(left == null || right == null){ if(left != null || right != null){ return false; } else { continue; } } int leftVal = left.val; int rightVal = right.val; if(leftVal != rightVal){ return false; } // 再把left和right子節點放進去，注意順序 // 順序必定是left的左節點和right的右節點，而後是left的右節點和right的左節點 queue.offer(left.left); queue.offer(right.right); queue.offer(left.right); queue.offer(right.left); } return true; } } 

總結

本篇文章講解了算法題目的思路和解法，代碼和筆記因爲純手打，不免會有紕漏，若是發現錯誤的地方，請第一時間告訴我，這將是我進步的一個很重要的環節。之後會按期更新算法題目以及各類開發知識點，若是您以爲寫得不錯，不妨點個關注，謝謝。