654. 最大二叉树
思路和由中后序列构建树的思路是一样的，只不过这里不是通过后序的最后一个数来进行分割，而是通过找最大值进行分割左右子树
迭代找出左子树和右子树的的边界，返回当前节点就行了
具体代码后续补
617. 合并二叉树
这里的return node不能到最后才去return，一定在某一个条件里面马上return了
/*** 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) {// console.log(root1,root2);// console.log(root1 === null)// return;//两个都为空if(root1 == null && root2 == null) return null;//root2为空let node = new TreeNode();if(root1 != null && root2 == null){let node = new TreeNode(root1.val);node.left = mergeTrees(root1.left, null);node.right = mergeTrees(root1.right, null);return node;}else if(root1 == null && root2 != null){let node = new TreeNode(root2.val);node.left = mergeTrees(null, root2.left);node.right = mergeTrees(null, root2.right);return node;}else{let node = new TreeNode(root1.val + root2.val);node.left = mergeTrees(root1.left, root2.left);node.right = mergeTrees(root1.right, root2.right);return node;}
};
700. 二叉搜索树中的搜索
/*** 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} val* @return {TreeNode}*/
var searchBST = function(root, val) {if(!root) return null;if(val > root.val){return searchBST(root.right, val);}else if(val < root.val){return searchBST(root.left, val);}else return root;};
98. 验证二叉搜索树
这道题有陷阱的，子树是BST，不代表这个树一定是BST
要清楚的特性是对于BST来说，起中序遍历的数组一定是一个升序数组
/*** 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 {boolean}*/
var isValidBST = function(root) {let inorder = [];inOrder(root, inorder);for(let i = 1; i < inorder.length; i++){if(inorder[i - 1] >= inorder[i]) return false;}return true;};function inOrder(root, inorder){if(!root) return;inOrder(root.left, inorder);inorder.push(root.val);inOrder(root.right, inorder);
}
)
