617. 合并二叉树
/**
* 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} root1
* @param {TreeNode} root2
* @return {TreeNode}
*/
var mergeTrees = function(root1, root2) {
if (root1 == null && root2 == null) return null;
if (root1 == null && root2 !== null) return root2;
if (root1 !== null && root2 == null) return root1;
const dfs = (root1, root2) => {
if (root1 == null && root2 == null) return null;
if (root1 == null && root2 !== null) return root2;
if (root1 !== null && root2 == null) return root1;
const root = new TreeNode(root1.val + root2.val);
root.left = dfs(root1.left, root2.left);
root.right = dfs(root1.right, root2.right);
return root;
}
return dfs(root1, root2);
};
637. 二叉树的层平均值
/**
* 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 {number[]}
*/
/* BFS 广度优先遍历 */
var averageOfLevels = function(root) {
if (root == null) return [];
const queue = [root];
const res = [];
while (queue.length) {
let curArr = [];
const length = queue.length;
for (let index = 0; index < length; index++) {
const node = queue.shift();
curArr.push(node.val);
if (node.left) {
queue.push(node.left);
}
if (node.right) {
queue.push(node.right);
}
}
/* 取总数 */
let sum = curArr.reduce((prev, current) => current + prev);
/* 计算平均值 */
res.push(sum / curArr.length);
}
return res;
};
剑指 Offer 27. 二叉树的镜像
简单
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {TreeNode}
*/
var mirrorTree = function(root) {
if (root == null) return null;
const dfs = (root) => {
if (root == null) return null;
const node = new TreeNode(root.val);
node.left = dfs(root.right);
node.right = dfs(root.left);
return node;
}
return dfs(root);
};
剑指 Offer 28. 对称的二叉树
简单
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
var isSymmetric = function(root) {
if(root == null) return true;
const help = (left, right) => {
if(left == null && right == null) return true;
if (left == null || right == null || left.val !== right.val) {
return false;
}
return help(left.left, right.right) && help(left.right, right.left);
}
return help(root.left, root.right);
};
剑指 Offer 32 - II. 从上到下打印二叉树 II
简单
剑指 Offer 32 - II. 从上到下打印二叉树 II
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number[][]}
*/
/*
BFS
*/
var levelOrder = function(root) {
if (root == null) return [];
const queue = [root];
const res = [];
while (queue.length) {
const length = queue.length;
let arr = [];
for (let index = 0; index < length; index++) {
const node = queue.shift();
arr.push(node.val);
if (node.left) {
queue.push(node.left);
}
if (node.right) {
queue.push(node.right);
}
}
res.push(arr);
}
return res;
};
剑指 Offer 54. 二叉搜索树的第k大节点
简单
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @param {number} k
* @return {number}
*/
var kthLargest = function(root, k) {
if (root == null) return 0;
let arr = [];
const dfs = (root) => {
if (root == null) return;
dfs(root.left);
arr.push(root.val);
dfs(root.right);
}
dfs(root);
arr.reverse();
return arr[k - 1];
};
剑指 Offer 55 - II. 平衡二叉树
简单
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
// 左右子树的深度不超过1，就是平衡二叉树
var isBalanced = function(root) {
if (root==null) return true;
const getHeight = (node) => {
if (node== null) return 0;
return Math.max(getHeight(node.left), getHeight(node.right)) + 1;
}
return Math.abs(getHeight(root.left) - getHeight(root.right)) <= 1 && isBalanced(root.left) && isBalanced(root.right);
};
剑指 Offer 68 - I. 二叉搜索树的最近公共祖先
简单
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @param {TreeNode} p
* @param {TreeNode} q
* @return {TreeNode}
*/
/*
经典题目
*/
var lowestCommonAncestor = function(root, p, q) {
if (root == null) return null;
const dfs = (root) => {
if (root == null) return null;
if (root.val == p.val || root.val == q.val) {
return root;
}
let left = dfs(root.left);
let right = dfs(root.right);
if (left && right) {
return root;
}
if (left == null) {
return right;
}
return left;
}
return dfs(root);
};
/* 这个方式真不错 */
var lowestCommonAncestor = function(root, p, q){
if(p.val < root.val && q.val < root.val){
return lowestCommonAncestor(root.left, p, q);
}else if(p.val > root.val && q.val > root.val){
return lowestCommonAncestor(root.right, p, q);
}else return root;
}
剑指 Offer 68 - II. 二叉树的最近公共祖先
简单
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @param {TreeNode} p
* @param {TreeNode} q
* @return {TreeNode}
*/
var lowestCommonAncestor = function(root, p, q) {
if (root == null) return null;
const dfs = (root) => {
if (root == null) return null;
if (root.val == p.val || root.val == q.val) {
return root;
}
let left = dfs(root.left);
let right = dfs(root.right);
if (left && right) {
return root;
}
if (left == null) {
return right;
}
return left;
}
return dfs(root);
};
剑指 Offer II 052. 展平二叉搜索树
简单
/**
* 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 {TreeNode}
*/
/*
先转成数组
在一个个变成新的搜索树
*/
var increasingBST = function(root) {
if (root == null) return null;
let prev = new TreeNode(-1);
let res = [];
const dfs = (root) => {
if (root == null) return;
dfs(root.left, prev.right);
res.push(root.val);
dfs(root.right, prev.right);
}
dfs(root);
/* 指针 */
let cur = prev;
for (const iterator of res) {
cur.right = new TreeNode(iterator);
cur = cur.right;
}
return prev.right;
};
剑指 Offer II 056. 二叉搜索树中两个节点之和
/**
* 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
* @param {number} k
* @return {boolean}
*/
/*
中序遍历 + 双指针
*/
var findTarget = function(root, k) {
let arr = [];
const inorder = (root) => {
if (root == null) return;
inorder(root.left);
arr.push(root.val);
inorder(root.right);
}
inorder(root);
let left = 0;
let right = arr.length - 1;
while (left < right) {
if (arr[left] + arr[right] > k) {
right--;
} else if (arr[left] + arr[right] < k) {
left++;
} else {
return true;
}
}
return false;
};
面试题 04.02. 最小高度树
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {number[]} nums
* @return {TreeNode}
*/
var sortedArrayToBST = function(nums) {
if (!nums.length) return null;
let i = Math.floor(nums.length / 2);
const root = new TreeNode(nums[i]);
root.left = sortedArrayToBST(nums.slice(0, i));
root.right = sortedArrayToBST(nums.slice(i + 1));
return root;
};
面试题 04.04. 检查平衡性
简单
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
var isBalanced = function(root) {
if (root == null) return true;
const help = (root) => {
if(root == null) return 0;
let left = help(root.left);
let right = help(root.right);
return Math.max(left, right) + 1;
}
return Math.abs(help(root.left) - help(root.right)) <=1 && isBalanced(root.left) && isBalanced(root.right)
};
面试题 17.12. BiNode
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {TreeNode}
*/
/*
这个题目有两个解决方案
1， 用中序遍历，获取值的数组，然后再创建一个新的树
2， 在原来的树上进行操作，使用指针的形式
*/
var convertBiNode = function(root) {
if (root == null) return null;
let prev= new TreeNode(-1);
let res = prev;
const inorder = (root) => {
if (root == null) return null;
inorder(root.left);
root.left = null;
prev.right = root;
/* 指针指向下一位 */
prev = root;
inorder(root.right);
}
inorder(root);
return res.right;
};
