算法笔记(二叉树1)

已于 2025-02-21 11:23:26 修改
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分类专栏： 数据结构 文章标签： 数据结构
于 2024-06-18 15:53:57 首次发布
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本文链接：https://blog.youkuaiyun.com/weixin_64704029/article/details/139775870
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leetcode144 二叉树的前序遍历

递归版本

public List<Integer> preorderTraversal(TreeNode root) {
    List<Integer> res = new ArrayList<>();
    preorder(root, res);
    return res;
}

public void preorder(TreeNode root, List<Integer> res) {
    if (root == null) {
        return;
    }
    res.add(root.val);
    preorder(root.left, res);
    preorder(root.right, res);
}

非递归版本:利用栈模拟

public List<Integer> preorderTraversal(TreeNode root) {
    List<Integer> res = new ArrayList<>();
    //利用栈模拟非递归遍历
    Stack<TreeNode> stack = new Stack<>();
    if (root == null) {
        return res;
    }
    stack.push(root);
    while (!stack.isEmpty()) {
        TreeNode pop = stack.pop();
        res.add(pop.val);
        if (pop.right != null) {
            stack.push(pop.right);
        }
        if (pop.left != null) {
            stack.push(pop.left);
        }
    }
    return res;
}

leetcode94 二叉树的中序遍历

 List<Integer> list = new ArrayList<>();

public List<Integer> inorderTraversal(TreeNode root) {
    mid(root);
    return list;
}

private void mid(TreeNode root) {
    if (root == null) return;
    mid(root.left);
    list.add(root.val);
    mid(root.right);
}

非递归版本

public List<Integer> inorderTraversal(TreeNode root) {
    List<Integer> result = new ArrayList<>();
    if (root == null) {
        return result;
    }
    Stack<TreeNode> stack = new Stack<>();
    TreeNode cur = root;
    while (cur != null || !stack.isEmpty()) {
        if (cur != null) {
            stack.push(cur);
            cur = cur.left;
        } else {
            cur = stack.pop();
            result.add(cur.val);
            cur = cur.right;
        }
    }
    return result;
}

leetcode145 二叉树的后序遍历

public List<Integer> postorderTraversal(TreeNode root) {
    List<Integer> list = new ArrayList();
    if (root == null) {
        return list;
    }
    postorder(root, list);
    return list;
}

private void postorder(TreeNode root, List<Integer> list) {
    if (root == null) {
        return;
    }
    postorder(root.left, list);
    postorder(root.right, list);
    list.add(root.val);
}

非递归版本

public List<Integer> postorderTraversal(TreeNode root) {
    List<Integer> res = new ArrayList<>();
    if (root == null) {
        return res;
    }
    Stack<TreeNode> stack = new Stack<>();
    stack.push(root);
    while (!stack.isEmpty()) {
        TreeNode node = stack.pop();
        res.add(node.val);
        if (node.left != null) {
            stack.push(node.left);
        }
        if (node.right != null) {
            stack.push(node.right);
        }
    }
    Collections.reverse(res);
    return res;
}

leetcode102 二叉树的层序遍历

public List<List<Integer>> levelOrder(TreeNode root) {
    List<List<Integer>> res = new ArrayList<>();
    Queue<TreeNode> queue = new LinkedList<>();
    if (root == null) {
        return res;
    }
    queue.add(root);
    while (!queue.isEmpty()) {
        int size = queue.size();
        List<Integer> temp = new ArrayList<>();
        for (int i = 0; i < size; i++) {
            TreeNode node = queue.poll();
            temp.add(node.val);
            if (node.left != null) {
                queue.offer(node.left);
            }
            if (node.right != null) {
                queue.offer(node.right);
            }
        }
        res.add(temp);
    }
    return res;
}

leetcode199 二叉树的右视图

在层序遍历中只加入该层的最后一个节点

public List<Integer> rightSideView(TreeNode root) {
    List<Integer> res = new ArrayList<>();
    Queue<TreeNode> queue = new LinkedList<>();
    if (root == null) {
        return res;
    }
    queue.add(root);
    while (!queue.isEmpty()) {
        int size = queue.size();
        for (int i = 0; i < size; i++) {
            TreeNode poll = queue.poll();
            if (poll.left != null) {
                queue.add(poll.left);
            }
            if (poll.right != null) {
                queue.add(poll.right);
            }
            if (i == size - 1) {
                res.add(poll.val);
            }
        }
    }
    return res;
}

leetcode637 二叉树层的平均值

public List<Double> averageOfLevels(TreeNode root) {
    List<Double> res = new ArrayList<>();
    Queue<TreeNode> queue = new LinkedList<>();
    if (root == null) {
        return res;
    }
    queue.add(root);
    while (!queue.isEmpty()) {
        double temp = 0;
        int size = queue.size();
        for (int i = 0; i < size; i++) {
            TreeNode node = queue.poll();
            temp += node.val;
            if (node.left != null) {
                queue.add(node.left);
            }
            if (node.right != null) {
                queue.add(node.right);
            }
        }
        temp = temp / size;
        res.add(temp);
    }
    return res;
}

leetcode429 N叉树的层序遍历

public List<List<Integer>> levelOrder(Node root) {
    List<List<Integer>> res = new ArrayList<>();
    Queue<Node> queue = new LinkedList<>();
    if (root == null) {
        return res;
    }
    queue.add(root);
    while (!queue.isEmpty()) {
        int size = queue.size();
        List<Integer> temp = new ArrayList<>();
        for (int i = 0; i < size; i++) {
            Node poll = queue.poll();
            temp.add(poll.val);
            List<Node> children = poll.children;
            for (int j = 0; j < children.size(); j++) {
                if (children.get(j) != null) {
                    queue.add(children.get(j));
                }
            }
        }
        res.add(temp);
    }
    return res;
}

leetcode515 在每个树行中的最大值

public List<Integer> largestValues(TreeNode root) {
    List<Integer> res = new ArrayList<>();
    if (root == null) {
        return res;
    }
    Queue<TreeNode> queue = new LinkedList<>();
    queue.add(root);
    while (!queue.isEmpty()) {
        int size = queue.size();
        int temp = Integer.MIN_VALUE;
        for (int i = 0; i < size; i++) {
            TreeNode poll = queue.poll();
            temp = Math.max(temp, poll.val);
            if (poll.left != null) {
                queue.add(poll.left);
            }
            if (poll.right != null) {
                queue.add(poll.right);
            }
        }
        res.add(temp);
    }
    return res;
}

leetcode116 填充每个节点的右侧指针

public Node connect(Node root) {
    Queue<Node> queue = new LinkedList<>();
    if (root == null) {
        return null;
    }
    queue.add(root);
    while (!queue.isEmpty()) {
        int size = queue.size();
        Node poll = queue.poll();
        if (poll.left != null) {
            queue.add(poll.left);
        }
        if (poll.right != null) {
            queue.add(poll.right);
        }
        for (int i = 1; i < size; i++) {
            Node cur = queue.poll();
            if (cur.left != null) {
                queue.add(cur.left);
            }
            if (cur.right != null) {
                queue.add(cur.right);
            }
            poll.next = cur;
            poll = cur;
        }
    }
    return root;
}

leetcode104 二叉树的最大深度

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

leetcode226 翻转二叉树

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

leetcode101 对称二叉树

public boolean isSymmetric(TreeNode root) {
    if (root == null) {
        return true;
    }
    return check(root.left, root.right);
}

public boolean check(TreeNode left, TreeNode right) {
    if (left == null && right != null) {
        return false;
    } else if (left != null && right == null) {
        return false;
    } else if (left == null && right == null) {
        return true;
    } else if (left.val != right.val) {
        return false;
    }

    //比较外侧节点
    boolean bool1 = check(left.left, right.right);
    //比较内侧节点
    boolean bool2 = check(left.right, right.left);
    return bool1 && bool2;
}

leetcode105 前序和中序构建二叉树

class Solution {
    Map<Integer, Integer> map = new HashMap<>();

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        for (int i = 0; i < preorder.length; i++) {
            map.put(inorder[i], i);
        }
        return find(preorder, 0, preorder.length, inorder, 0, inorder.length);
    }

    private TreeNode find(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd) {
        if (preStart >= preEnd || inStart >= inEnd) {
            return null;
        }
        //构建根节点
        int rootIndex = map.get(preorder[preStart]);
        TreeNode root = new TreeNode(inorder[rootIndex]);
        int len = rootIndex - inStart;//保存中序遍历中左子树的个数；
        //切分前序数组
        root.left = find(preorder, preStart + 1, preStart + len + 1, inorder, inStart, inEnd);
        root.right = find(preorder, preStart + len + 1, preEnd, inorder, rootIndex + 1, inEnd);
        return root;
    }
}

leetcode106 中序和后序构建二叉树

class Solution {
    Map<Integer,Integer> map ;
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        map = new HashMap<>();
        for (int i = 0; i < inorder.length; i++) {
            map.put(inorder[i],i);//map用来存储元素的位置
        }
        return createNode(inorder,0, inorder.length, postorder,0, postorder.length);
    }

    public TreeNode createNode(int[] inorder,int inBegin,int inEnd,int[] postorder,
                               int postBegin,int postEnd){
        if (inBegin >= inEnd || postBegin >= postEnd){
            return null;
        }
        int rootIndex = map.get(postorder[postEnd - 1]);//根节点位置
        TreeNode root = new TreeNode(inorder[rootIndex]);//获取根节点在中序中的位置
        int left = rootIndex - inBegin;//保存左子树的个数,可以计算出右子树的个数
        root.left = createNode(inorder,inBegin,rootIndex,
                postorder,postBegin,postBegin+left);
        root.right = createNode(inorder,rootIndex+1,inEnd,
                postorder,postBegin+left,postEnd-1);
        return root;
    }
}

leetcode108 将有序数组转化为二叉搜索树

二分搜索的思路

public TreeNode sortedArrayToBST(int[] nums) {
    if (nums == null || nums.length == 0){
      return null;
    }
    return set(nums,0, nums.length-1);
  }
  private TreeNode set(int[] nums,int left,int right){
    if (left > right){
      return null;
    }
    int index=(left+right)>>>1;
    TreeNode root = new TreeNode(nums[index]);
    root.left = set(nums,left,index - 1);
    root.right = set(nums,index+1,right);
    return root;
  }

leetcode 98验证二叉搜索树

中序遍历+判断

public boolean isValidBST(TreeNode root) {
    List<Integer> list = new ArrayList<>();
    mid(root,list);
    for (int i = 1; i < list.size(); i++) {
      if (list.get(i)<list.get(i-1)){
        return false;
      }
    }
    return true;
  }

  private void mid(TreeNode root,List<Integer> list) {
    if (root == null){
      return;
    }
    mid(root.left,list);
    list.add(root.val);
    mid(root.right,list);
  }

leetcode114 二叉树展开为链表

class Solution {
    public void flatten(TreeNode root) {
        //思路：将根节点的左子树挂接到右子树上，将原先的右子树挂接到新的右子树的后面
        while (root != null) {
            if (root.left == null) {
                root = root.right;
            } else {
                TreeNode pre = root.left;
                while (pre.right != null) {
                    pre = pre.right;//找到左子树最右边的结点
                }
                pre.right = root.right;//将原先右子树挂接
                root.right = root.left;
                root.left = null;
                root = root.right;
            }
        }
    }
}

leetcode 112 路径总和

给你二叉树的根节点 root 和一个表示目标和的整数 targetSum 。判断该树中是否存在 根节点到叶子节点 的路径，这条路径上所有节点值相加等于目标和 targetSum 。如果存在，返回 true ；否则，返回 false 。

public boolean hasPathSum(TreeNode root, int targetSum) {
        if (root == null){
            return false;
        }
        //回溯过程
        targetSum -= root.val;
        if (root.left == null && root.right == null){
            return targetSum == 0;
        }
        if (root.left!=null){
            boolean left = hasPathSum(root.left,targetSum);
            if (left){
                return true;
            }
        }
        if (root.right != null){
            boolean right = hasPathSum(root.right,targetSum);
            if (right){
                return true;
            }
        }
        return false;
    }

leetcode 437 路径总和三

给定一个二叉树的根节点 root ，和一个整数 targetSum ，求该二叉树里节点值之和等于 targetSum 的 路径 的数目。路径 不需要从根节点开始，也不需要在叶子节点结束，但是路径方向必须是向下的（只能从父节点到子节点）。
思路：dfs+递归

int count = 0;
public int pathSum(TreeNode root, int targetSum) {
  if (root == null){
    return 0;
  }
  dfs(root,targetSum);
  pathSum(root.left,targetSum);
  pathSum(root.right,targetSum);
  return count;
}

private void dfs(TreeNode root, int targetSum) {
  if (root == null){
    return;
  }
  if (root.val == targetSum){
    count++;
  }
  dfs(root.left,targetSum-root.val);
  dfs(root.right,targetSum-root.val);
}

leetcode 236 二叉树的最近公共祖先

 public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
     if (root == null || root == p || root == q){
       return root;
     }
     TreeNode left = lowestCommonAncestor(root.left,p,q);
     TreeNode right = lowestCommonAncestor(root.right,p,q);
    if (left == null && right != null) {
      return right;
    } else if (left != null && right == null) {
      return left;
    } else if (left == null && right == null) {
      return null;
    }else {
      return root;
    }
  }

leetcode 124 二叉树的最大路径和

int ans = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        dfs(root);
        return ans;
    }

    int dfs(TreeNode root){
        // 计算以当前节点为最高点的时候的最大路径和
        if(root == null){
            return 0;
        }
        int left = dfs(root.left);
        int right = dfs(root.right);
        int t = root.val;
        if(left > 0){
            t += left;
        }
        if(right > 0){
            t += right;
        }
        ans = Math.max(ans,t);
        return Math.max(root.val , Math.max(left,right) + root.val);
    }

leetcode 257 二叉树的所有路径

public List<String> binaryTreePaths(TreeNode root) {
    List<String> res = new ArrayList<>();
    if (root == null ){
      return res;
    }
    dfs1(root,res,"");
    return res;
  }

  private void dfs1(TreeNode root, List<String> res, String s) {
    if (root == null){
      return;
    }
    s += root.val;
    if (root.left == root.right){
      res.add(s);
      return;
    }
    s+="->";
    dfs1(root.left,res,s);
    dfs1(root.right,res,s);
  }
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