LeetCode刷题 Day18
513. Find Bottom Left Tree Value
Given the root of a binary tree, return the leftmost value in the last row of the tree.
Example 1:
Input: root = [2,1,3]
Output: 1
Example 2:
Input: root = [1,2,3,4,null,5,6,null,null,7]
Output: 7
BFS法:
- 每当在i === 0的时候更新leftResult
代码:
var findBottomLeftValue = function(root) {
let queue = [];
let leftResult = -1;
queue.push(root);
while (queue.length > 0) {
const len = queue.length;
for (let i = 0; i < len; i++) {
const node = queue.shift();
if (i === 0) leftResult = node.val;
if (node.left) {
queue.push(node.left);
}
if (node.right) {
queue.push(node.right);
}
}
}
return leftResult
};
DFS法:
- 这个要理解遍历的顺序，当在遍历到末尾进行回溯的时候, 必然是先回到左边的节点 再去右边的节点，再结合最大的depth去判断最后一行最左边节点
代码:
var findBottomLeftValue = function(root) {
let maxDepth = -1;
let leftMostValue = -1;
var helper = function(root, depth) {
if (!root) return;
if (!root.left && !root.right) {
if (maxDepth < depth) {
maxDepth = depth;
leftMostValue = root.val;
}
return;
}
helper(root.left, depth + 1);
helper(root.right, depth + 1);
}
helper(root, 0);
return leftMostValue;
};
时间复杂度: O(n), 空间复杂度: O(n)
112. Path Sum
Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.
A leaf is a node with no children.
Example 1:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
Explanation: The root-to-leaf path with the target sum is shown.
Example 2:
Input: root = [1,2,3], targetSum = 5
Output: false
Explanation: There two root-to-leaf paths in the tree:
(1 --> 2): The sum is 3.
(1 --> 3): The sum is 4.
There is no root-to-leaf path with sum = 5.
Example 3:
Input: root = [], targetSum = 0
Output: false
Explanation: Since the tree is empty, there are no root-to-leaf paths.
思路:
- 这道题的重点在于 !root的时候应该false, 根据 example 3的描述
- 用“隐藏式”的回溯比较方便
代码:
var hasPathSum = function(root, targetSum) {
if (!root) return false;
if (!root.left && !root.right && targetSum === root.val) return true;
return hasPathSum(root.left, targetSum - root.val) || hasPathSum(root.right, targetSum - root.val)
};
时间复杂度: O(n),还是把节点都遍历一遍 空间复杂度: O(n)
113. Path Sum II
Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum . Each path should be returned as a list of the node values, not node references.
A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.
Example 1:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: [[5,4,11,2],[5,8,4,5]]
Explanation: There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22
Example 2:
Input: root = [1,2,3], targetSum = 5
Output: []
Example 3:
Input: root = [1,2], targetSum = 0
Output: []
思路:
- 这个和 Path SumI是一样的，但是要记录path, 所以将path作为参数放入递归中, 当符合条件的时候就res.push([...path])
代码:
var pathSum = function(root, targetSum) {
let res = [];
var helper = function(root, path, targetSum) {
if (!root) return;
if (!root.left && !root.right) {
if (root.val === targetSum) {
path.push(root.val);
res.push([...path]);
path.pop();
return;
}
}
path.push(root.val);
helper(root.left, path, targetSum - root.val);
helper(root.right, path, targetSum - root.val);
path.pop();
}
helper(root, [], targetSum);
return res;
};
时间复杂度: O(n2), 这里要考虑deep copy的问题，复杂度是O(n),再结合树本身的遍历O(n)。最后应该是O(n2) 空间复杂度: O(n)
105. Construct Binary Tree from Preorder and Inorder Traversal
Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.
Example 1:
Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]
Example 2:
Input: preorder = [-1], inorder = [-1]
Output: [-1]
思路:
- 有点像二分搜索, 但是left和right是用来定位inorder的左右区间
- prorder可以以从左到右的顺序获得根节点的value，通过根节点的value在inorder中进行定位，将二叉树分为左右两支。 再在左右两支中继续相同逻辑，直到left > right，return null
代码:
var buildTree = function(preorder, inorder) {
var helper = function (left, right) {
if (left > right) return null;
const rootVal = preorder.shift();
const root = new TreeNode(rootVal);
root.left = helper(left, inorder.indexOf(rootVal) - 1);
root.right = helper(inorder.indexOf(rootVal) + 1, right);
return root;
}
return helper(0, inorder.length - 1);
};
时间复杂度: O(n) 空间复杂度: O(n)
106. Construct Binary Tree from Inorder and Postorder Traversal
Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.
Example 1:
Input: inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
Output: [3,9,20,null,null,15,7]
Example 2:
Input: inorder = [-1], postorder = [-1]
Output: [-1]
思路:
- 和105题一样， 不同的是要注意后序遍历的顺序和先序遍历是反的。
var buildTree = function(inorder, postorder) {
var helper = function(left, right) {
if (left > right) return null;
const rootVal = postorder.pop();
const root = new TreeNode(rootVal);
root.right = helper(inorder.indexOf(rootVal) + 1, right);
root.left = helper(left, inorder.indexOf(rootVal) - 1);
return root;
}
return helper(0, inorder.length - 1);
};
时间复杂度: O(n) 空间复杂度: O(n)
