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前言

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二叉树的递归分为「遍历」和「分解问题」两种思维模式，这道题需要用到「遍历」的思维。 代码看起来虽然多，但思路非常简单：用 path 维护递归遍历的路径，到达叶子节点的时候判断字典序最小的路径。 不要忘了在叶子节点的时候也要正确维护 path 变量，而且要把 StringBuilder 中的字符串反转才是目想要的答案。

一、力扣298. 二叉树最长连续序列

/*** 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 {int res = 1;public int longestConsecutive(TreeNode root) {fun(root, 1, Integer.MIN_VALUE);return res;}public void fun(TreeNode root, int len, int parentVal){if(root == null){return ;}if(root.val == parentVal+1){len ++;}else{len = 1;}res = Math.max(len, res);fun(root.left, len, root.val);fun(root.right,len,root.val);}
}

二、力扣988. 从叶结点开始的最小字符串

/*** 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 {String res = null;StringBuilder sb = new StringBuilder();public String smallestFromLeaf(TreeNode root) {fun(root);return res;}public void fun(TreeNode root){if(root == null){return;}sb.append((char)('a'+root.val));if(root.left == null && root.right == null){sb.reverse();String s = sb.toString();if(res == null || res.compareTo(s) > 0){res = s;}sb.reverse();sb.deleteCharAt(sb.length()-1);return;}fun(root.left);fun(root.right);sb.deleteCharAt(sb.length()-1);}
}

三、力扣1022. 从根到叶的二进制数之和

/*** 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 {int res = 0;int path = 0;public int sumRootToLeaf(TreeNode root) {fun(root);return res;}public void fun(TreeNode root){if(root == null){return;}path = path << 1 | root.val;if(root.left == null && root.right == null){res += path;}fun(root.left);fun(root.right);path = path >> 1;}
}

四、力扣1457. 二叉树中的伪回文路径

/*** 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 {List<Integer> path = new ArrayList<>();int[] arr = new int[10];int count = 0;public int pseudoPalindromicPaths (TreeNode root) {fun(root);return count;}public void fun(TreeNode root){if(root == null){return ;}path.add(root.val);arr[root.val] ++;if(root.left == null && root.right == null){int flag = 0;for(int i = 0; i < 10; i ++){if(arr[i]%2 == 1){flag ++;}}if(flag <= 1){count ++;}arr[root.val] --;return;}fun(root.left);fun(root.right);path.remove(path.size()-1);arr[root.val] --;}
}

五、力扣