zoukankan      html  css  js  c++  java
  • Leetcode 623 在二叉树中增加一行 BFS与DFS

       BFS 解法:

    private final void addRow(Stack<TreeNode> stack, int v) {
            while (!stack.empty()) {
                TreeNode node = stack.pop();
                TreeNode newNodeLeft = new TreeNode(v);
                TreeNode newNodeRight = new TreeNode(v);
                newNodeLeft.left = node.left;
                newNodeRight.right = node.right;
                node.left = newNodeLeft;
                node.right = newNodeRight;
            }
        }
    
        /**
         * @Author Niuxy
         * @Date 2020/7/17
         * @Description BFS
         */
        public final TreeNode addOneRow(TreeNode root, int v, int d) {
            if (d == 1) {
                TreeNode newRoot = new TreeNode(root.val);
                newRoot.left = root.left;
                newRoot.right = root.right;
                root.left = newRoot;
                root.right = null;
                root.val = v;
                return root;
            }
            Stack<TreeNode> stack1 = new Stack<TreeNode>();
            Stack<TreeNode> stack2 = new Stack<TreeNode>();
            stack1.push(root);
            int length = 1;
            while (!(stack1.empty() && stack1.empty())) {
                while (!stack1.empty()) {
                    boolean isInsert = isInsert(stack1, stack2, d, v, length, root);
                    if (isInsert) {
                        return root;
                    }
                }
                length++;
                while (!stack2.empty()) {
                    boolean isInsert = isInsert(stack2, stack1, d, v, length, root);
                    if (isInsert) {
                        return root;
                    }
                }
                length++;
            }
            return root;
        }
    
        private final boolean isInsert(Stack<TreeNode> stack1, Stack<TreeNode> stack2, int d,
                                       int v, int length, TreeNode root) {
            if (length == d - 1) {
                addRow(stack1, v);
                return true;
            }
            TreeNode node = stack1.pop();
            if (node.left != null) {
                stack2.push(node.left);
            }
            if (node.right != null) {
                stack2.push(node.right);
            }
            return false;
        }

      DFS解法:

    public final TreeNode addOneRowDFS(TreeNode root, int v, int d) {
            if (d == 1) {
                TreeNode newRoot = new TreeNode(root.val);
                newRoot.left = root.left;
                newRoot.right = root.right;
                root.left = newRoot;
                root.right = null;
                root.val = v;
                return root;
            }
            Stack<TreeNode> stack = new Stack<TreeNode>();
            dfs(root, v, d, stack);
            addRow(stack, v);
            return root;
        }
    
        public final void dfs(TreeNode node, int v, int d, Stack<TreeNode> stack) {
            if (node == null) {
                return;
            }
            if (d == 2) {
                stack.push(node);
                return;
            }
            dfs(node.left, v, d - 1, stack);
            dfs(node.right, v, d - 1, stack);
        }
    
    
        private final void addRow(Stack<TreeNode> stack, int v) {
            while (!stack.empty()) {
                TreeNode node = stack.pop();
                TreeNode newNodeLeft = new TreeNode(v);
                TreeNode newNodeRight = new TreeNode(v);
                newNodeLeft.left = node.left;
                newNodeRight.right = node.right;
                node.left = newNodeLeft;
                node.right = newNodeRight;
            }
        }

  • 相关阅读:
    phone 接口实现
    linux 下查找文件或者内容常有命令【转】
    【转】线程问题引起的内存泄露
    Canvas的save和restore 之二
    getIdentifer()函数的用法
    canvas.clipRect
    blob
    Android中Context
    canvas中的save用法之一
    sharedPreferences
  • 原文地址:https://www.cnblogs.com/niuyourou/p/13329190.html
Copyright © 2011-2022 走看看