Leetcode （力扣）做题记录 hot100（42，104，226，101）

力扣第42题：接雨水

42. 接雨水 - 力扣（LeetCode）

左边遍历一次记录左侧最大值 右边同理，最后遍历一次 左侧右侧最小值减去当前值即可。

class Solution {public int trap(int[] height) {int n = height.length;int[] leftMax = new int[n];int[] rightMax = new int[n];int sum = 0;leftMax[0] = height[0];for(int i = 1;i<n;i++ ){leftMax[i] = Math.max(leftMax[i - 1],height[i]);}rightMax[n -  1] = height[ n -1];for(int i = n -2;i>=0;i--){rightMax[i] = Math.max(rightMax[i+1],height[i]);}for(int i =0;i<n;i++){sum =sum + Math.min(leftMax[i],rightMax[i]) - height[i];}return sum;}
}

力扣第104题：二叉树的最大深度

104. 二叉树的最大深度 - 力扣（LeetCode）

/*** 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 {public int maxDepth(TreeNode root) {if(root == null){return 0;}else{int k = maxDepth(root.left);int l = maxDepth(root.right);return Math.max(l,k) +1;}}
}

力扣第226题：翻转二叉树

226. 翻转二叉树 - 力扣（LeetCode）

/*** 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 {public TreeNode invertTree(TreeNode root) {if(root == null){return null;}TreeNode left = invertTree(root.left);TreeNode right = invertTree(root.right);root.left = right;root.right = left;return root}
}

力扣第101题：对称二叉树

101. 对称二叉树 - 力扣（LeetCode）

/*** 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 {public boolean isSymmetric(TreeNode root) {return cheek(root.left,root.right);}private boolean cheek(TreeNode l,TreeNode r){if(l == null&& r == null){return true;}if(l == null|| r == null){return false;}return l.val==r.val && cheek(l.left,r.right) && cheek(l.right,r.left);}
}

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