hot100之二叉树下

凉砧掌 发表于 2025-6-19 20:21:21

二叉树的右视图(199)

class Solution {
List<Integer> res = new ArrayList<>();
public List<Integer> rightSideView(TreeNode root) {
   dfs(root, 0);
   return res;
}
private void dfs(TreeNode node, int depth){
   if (node == null) return;

   if (res.size() == depth){
         res.add(node.val);
   }

   dfs(node.right, depth+1);
   dfs(node.left, depth+1);
}
}

[*]分析
因为一层只需要一个节点, 以depth作为限制
二叉树展开为链表(114)

class Solution {
public void flatten(TreeNode root) {
   if (root == null) return;

   flatten(root.left);
   flatten(root.right);
   TreeNode lef = root.left;
   TreeNode rig = root.right;

   root.left = null;
   root.right = lef;
   TreeNode curr = root;

   while (curr.right != null){
         curr = curr.right;
   }curr.right = rig;

}
}

[*]分析
将左子树截断放到右端, 再将右子树放在左子树末尾
因为递归原因, 左子树必然为链表
从前序与中序遍历序列构造二叉树(105)

class Solution {
Map<Integer, Integer> map;
public TreeNode buildTree(int[] preorder, int[] inorder) {
   int n = preorder.length;
   map = new HashMap<>();
   for (int i = 0; i < n; i++){
         map.put(inorder, i);
   }

   return dfs(preorder, 0, n, 0, n);
}

private TreeNode dfs(int[] preorder, int lefPre, int rigPre, int lefIn, int rigIn){
   if (lefPre == rigPre) return null;

   int lefSize = map.get(preorder) - lefIn;
   TreeNode lefNode = dfs(preorder, lefPre+1, lefPre+1+lefSize, lefIn, lefIn+lefSize);
   TreeNode rigNode = dfs(preorder, lefPre+1+lefSize, rigPre, lefIn+1+lefSize, rigIn);
   return new TreeNode(preorder, lefNode, rigNode);
}
}

[*]分析
以前序遍历取根, 切分中序遍历的左右子树作构造
路径总和(437)

class Solution {
Map<Long, Integer> map = new HashMap<>();
int res;
public int pathSum(TreeNode root, int targetSum) {
   map.put(0L,1);
   dfs(root, targetSum, 0L);
   return res;
}

private void dfs(TreeNode node, int targetSum, long subSum){
   if (node == null) return;
   subSum += node.val;
   if (map.containsKey(subSum - targetSum)) res +=map.get(subSum - targetSum);
   map.put(subSum, map.getOrDefault(subSum, 0)+1);
   dfs(node.left, targetSum, subSum);
   dfs(node.right, targetSum, subSum);
   map.put(subSum, map.get(subSum)-1);
}
}

[*]分析
又是一个前缀和题目
二叉树的最近公共祖先(236)

class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
   if (root == null || p == root || q == root) return root;

   TreeNode lefNode = lowestCommonAncestor(root.left, p, q);
   TreeNode rigNode = lowestCommonAncestor(root.right, p, q);

   if (lefNode == null && rigNode == null) return null;
   if (lefNode != null && rigNode != null) return root;

   return lefNode != null ? lefNode : rigNode;

}
}

[*]分析
找到子节点向上return, 两节点未相遇则继续向上return

[*]感悟
二叉树题目要注意需要返回给父节点的元素
二叉树中的最大路径和(124)

class Solution {
int res = Integer.MIN_VALUE;
public int maxPathSum(TreeNode root) {
   dfs(root);
   return res;
}
private int dfs(TreeNode node){
   if (node == null) return 0;

   int lefMax = Math.max(dfs(node.left), 0);
   int rigMax = Math.max(dfs(node.right), 0);

   res = Math.max(res, lefMax+ node.val + rigMax);
   return Math.max(lefMax, rigMax) + node.val;
}
}

[*]分析
以单向链作返回父节点元素, 取左右链+节点与记录的最大值作对比

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hot100之二叉树下