114. Flatten Binary Tree to Linked List

[Tcdian]

114. Flatten Binary Tree to Linked List

给定一个二叉树，原地将它展开为链表。

例如，给定二叉树

    1
   / \
  2   5
 / \   \
3   4   6

将其展开为：

1
 \
  2
   \
    3
     \
      4
       \
        5
         \
          6

[Tcdian]

Solution 1

• JavaScript Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {void} Do not return anything, modify root in-place instead.
 */
var flatten = function(root) {
    if (root === null) {
        return;
    }
    flatten(root.left);
    flatten(root.right);
    const next = root.right;
    root.right = root.left;
    root.left = null;
    while (root.right !== null) {
        root = root.right;
    }
    root.right = next;
};

• TypeScript Solution

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

/**
 Do not return anything, modify root in-place instead.
 */
function flatten(root: TreeNode | null): void {
    if (root === null) {
        return;
    }
    flatten(root.left);
    flatten(root.right);
    const next = root.right;
    root.right = root.left;
    root.left = null;
    while (root.right !== null) {
        root = root.right;
    }
    root.right = next;
};

Solution 2

• JavaScript Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {void} Do not return anything, modify root in-place instead.
 */
var flatten = function(root) {
    let current = root;
    while(current !== null) {
        if (current.left !== null) {
            const next = current.left;
            let patrol = next;
            while(patrol.right !== null) {
                patrol = patrol.right;
            }
            patrol.right = current.right;
            current.right = next;
            current.left = null;
        }
        current = current.right;
    }
};

• TypeScript Solution

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

/**
 Do not return anything, modify root in-place instead.
 */
function flatten(root: TreeNode | null): void {
    if (root === null) {
        return;
    }
    flatten(root.left);
    flatten(root.right);
    let patrol = root.left;
    if (patrol !== null) {
        const next = root.right;
        root.right = patrol;
        while(patrol.right !== null) {
            patrol = patrol.right;
        }
        patrol.right = next;
        root.left = null;
    }
};